Physics

1. The focal length of a convex mirror is 10 cm and the object distance is 20 cm. Determine (a) the image distance (b) the magnification of image

Known :

The focal length (f) = -10 cm

The minus sign indicates that the focal point of convex mirror is virtual

The object distance (d_o) = 20 cm

Solution :

Formation of image by concave mirror :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/10 – 1/20 = -2/20 – 1/20 = -3/20

d_i = -20/3 = -6.7 cm

The minus sign indicates that the image is virtual.

The magnification of image :

m = – d_i / d_o = -(-6.7)/20 = 6.7/20 = 0.3

m = 0,3 time smaller than the object.

The plus sign indicates that the image is upward.

2. A 10-cm high object is placed in front of a convex mirror with focal length 20 cm. Determine the image height if the object distance is (a) 10 cm (b) 30 cm (c) 40 cm (d) 50 cm

Known :

The focal length of convex mirror (f) = -20 cm

The minus sign indicates that the focal point is virtual

The radius of curvature (r) = 2 f = 2(20) = 40 cm

The object height (h) = 10 cm

Solution :

a) the focal length (f) = -20 cm and the object distance (d_o) = 10 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/10 = -1/20 – 2/20 = -3/20

d_i = -20/3 = -6.7

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image (m) :

m = –d_i / d_o = -(-6.7)/10 = 6.7/10 = 0.67

The plus sign indicates that the image is upright.

The image is 0.67 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.67) = 6.7 cm

b) the focal length (f) = -20 cm and the object distance (d_o) = 30 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/30 = -3/60 – 2/60 = -5/60

d_i = -60/5 = -12

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image (m) :

m = –d_i / d_o = -(-12)/30 = 12/30 = 0.4

The plus sign indicates that the image is upright.

The image is 0,4 times smaller the object.

The height of image (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.4) = 4 cm

c) The focal length (f) = -20 cm and the object distance (d_o) = 40 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/40 = -2/40 – 1/40 = -3/40

d_i = -40/3 = -13.3

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

The magnification of image (m) :

m = – d_i / d_o = -(-13.3)/40 = 13.3/40 = 0.3

The plus sign indicates that the image is upright.

The image is 0.3 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.3) = 3 cm

d) The focal length (f) = -20 cm and the object distance (d_o) = 50 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/50 = -5/100 – 2/100 = -7/100

d_i = -100/7 = -14.3

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

The magnification of image (m) :

m = – d_i / d_o = -(-14.3)/50 = 14.3/50 = 0.3

The plus sign indicates that the image is upright.

The image is 0.3 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.3) = 3 cm

3. On object is 20 cm in front of convex mirror. If the image height is 1/5 times the object height, determine (a) the image length b) the focal length c) the properties of image

Known :

The object distance (d_o) = 20 cm

The image height (h_i) = 1/5 h = 0.2 h

The object height (h) = h

Solution :

a) the image distance (d_i)

Formula of the magnification of image :

m = h_i / h_o = 0.2h / h = 0.2

The plus sign indicates that the image is upright.

The image is 0.2 smaller than the object.

The image distance (d_i) :

–d_i = m d_o

d_i = – m d_o = -(0.2)(20 cm) = -4 cm

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

b) The focal length (f)

The focal length (f) :

1/f = 1/d_o + 1/ d_i = 1/20 – 1/4 = 1/20 – 5/20 = -4/20

f = -20/4 = -5 cm

The minus sign indicates that the focal point is virtual.

c) The properties of image :

– Upright

– Smaller

– virtual

4. Light strikes a convex mirror parallel to the axis will be reflected ….

A. towards the focal point of the mirror

B. as from the focal point of the mirror

C. through the center of the curvature of the mirror

D. perpendicular to the mirror plane

Solution

The problem drew in the figure below.

The correct answer is B.

5. A biker sees the image of a motorcycle behind it 1/6 times its original size when the distance between the biker and motorcycle is 30 meters. Determine the radius of curvature of the rear view mirror…

A. 7.14 m

B. 8.57 m

C. 12.00 m

D. 24.00 m

Known :

Magnification of image (M) = 1/6 times

Object distance (d) = 30 meter

Wanted: The radius of curvature of the rear view mirror (R)

Solution :

Calculate the distance of the image (d’)

Since the magnification of image (M) and the object distance (s) has been known, the image distance can be known using the formula of image magnification :

A negative sign means the image is virtual. The image is 5 meters behind the convex mirror.

Calculate the focal length (f)

Since the object distance (d) and the image distance (d’), then the focal length can be calculated using the formula of the mirror :

The radius of curvature (R)

The radius of curvature of a convex mirror is twice the focal length of a convex mirror.
R = 2 f = 2 (6 meters) = 12 meters
The radius of curvature of the convex mirror is 12 meters.
The correct answer is C.

6. The convex mirror is chosen as the rearview mirror of a motorcycle because the properties of the image produced by the mirror are…

A. real, upright, minimized

B. real, upright, enlarged

C. virtual, upright, minimized

D. virtual, upright, enlarged

Solution :

Based on the two figure above can concluded that the properties of the image are virtual, upright, minimized.

The correct answer is C.

7. An object is 12 cm in front of a convex mirror with a radius of 6 cm. The properties of the image are…

A. real, inverted at a distance of 12 cm

B. real, upright at a distance of 4 cm

C. virtual, upright at a distance of 2.4 cm

D. virtual, inverted at a distance of 6 cm

Known :

Object’s distance (d) = 12 cm

Radius of the convex mirror (r) = 6 cm.

The focal length of the convex mirror (f) = 6 cm / 2 = -3 cm

The focal length of a convex mirror is signed negative because it is virtual. Virtual because it is not passed by light.

Wanted : Properties of image

Solution :

Distance of image (d’) :

1/d’ = 1/f – 1/d = -1/3 – 1/12 = -4/12 – 1/12 = -5/12

d’ = -12/5 = -2.4 cm

The image distance signed negative means image is virtual.

Magnification of image (m):

m = -d’ / d = -(-2.4) / 12 = 2.4 / 12 = 0.2 times

The magnification of image signed positive means that the image is upright and the magnification of image is 0.2 means the image size is smaller than the object size (reduced).

The correct answer is C.

[wpdm_package id=’860′]

1. Concave mirror problems and solutions
2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
10. Optical instrument telescopes problems and solutions

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1. An object is placed 10 cm from a concave mirror. The focal length is 5 cm. Determine (a) The image distance (b) the magnification of image

Known :

The focal length (f) = 5 cm

The object distance (d_o) = 10 cm

Solution :

Formation of image by concave mirror :

The image distance :

1/d_i = 1/f – 1/d_o = 1/5 – 1/10 = 2/10 – 1/10 = 1/10

d_i = 10/1 = 10 cm

The image distance is 10 cm.

The magnification :

m = –d_i / d_o = -10/10 = -1

1 means that the image is the same as the object.

The minus sign indicates that the image is inverted. If the sign is positive than the image is upright.

2. A 5-cm-high object is placed in front of a concave mirror with a radius of curvature of 20 cm. Determine the image height if the object distance is 5 cm, 15 cm, 20 cm, 30 cm.

Known :

The radius of curvature (r) = 20 cm

The focal length (f) = R/2 = 20/2 = 10 cm

The object height (h_o) = 5 cm

Solution :

a) the focal length (f) = 10 cm and the object distance (d_o) = 5 cm

Formation of image by concave mirror :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/10 – 1/5 = 1/10 – 2/10 = -1/10

d_i = -10/1 = -10 cm

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image (m) :

m = –d_i / d_o = -(-10)/5 = 10/5 = 2

The plus sign indicates that the image is upright.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (5 cm)(2) = 10 cm

The image height is 10 cm.

b) The focal length (f) = 10 cm and the object distance (d_o) = 15 cm

Formation of image by concave mirror :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/10 – 1/15 = 3/30 – 2/30 = 1/30

d_i = 30/1 = 30 cm

The plus sign indicates that the image is real or the image is 30 cm in front of the mirror, on the same side as the object.

The magnification of image (m) :

m = –d_i / d_o = -30/15 = -2

The minus sign indicates that the image is inverted.

The image is 2 times larger than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (5 cm)(2) = 10 cm

The image height is 10 cm.

c) The focal length (f) = 10 cm and the object distance (d_o) = 20 cm

Formation of image by concave mirror :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/10 – 1/20 = 2/20 – 1/20 = 1/20

d_i = 20/1 = 20 cm

The positive sign indicates that the image is real or the image is 20 cm in front of the mirror, on the same side as the object.

The magnification of image (m) :

m = –d_i / d_o = -20/20 = -1

The negative sign means the image is inverted.

The image height (hi) :

m = h_i / h_o

h_i = h m = (5 cm)(1) = 5 cm

d) The focal length (f) = 10 cm and the object distance (d_o) = 30 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/10 – 1/30 = 3/30 – 1/30 = 2/30

d_i = 30/2 = 15 cm

The plus sign indicates that the image is real or the image is 15 cm in front of the mirror, on the same side as the object.

The magnification of image (m) :

m = –d_i / d_o = -15/30 = -0.5

The minus sign indicates that the image is inverted.

The image is 0.5 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (5 cm)(0.5) = 2.5 cm

3. An image an by a concave mirror is 4 times greater than the object. If the radius of curvature 20 cm, determine the object distance in front of the mirror!

Known :

The magnification of image (m) = 4

The radius of curvature (r) = 20 cm

The focal length (f) = r/2 = 20/2 = 10 cm

Wanted : The object distance (d_o)

Solution :

m = – d_i / d_o

4 = – d_{i /} d_o

– d_i = 4 d_o

d_i = – 4 d_o

1/f = 1/d_o + 1/d_i

1/10 = 1/d_o + 1/4d_o

4/40 = 4/4d_o + 1/4d_o

4/40 = 5/4d_o

(4)(4s) = (5)(40)

16 d_o = 200

d_o = 12.5 cm

The object distance = 12.5 cm.

4. A 1-cm high object is placed 10 cm from a concave mirror with the focal length, f = 15 cm. Determine :

A. The image distance ?

B. The image height?

C. The properties of image formed by the concave mirror?

Known :

The object height (h) = 1 cm

The object distance (d_o) = 10 cm

The focal length of the concave mirror (f) = 15 cm

Solution :

A. The image distance (d_i)

1/f = 1/d_o + 1/d_i

1/d_i = 1/f – 1/d_o = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

d_i = -30/1 = -30 cm

The negative sign indicates that the image is virtual or the image is behind the mirror.

B. The image height (h_i)

The magnification of the image (M) :

M = -d_i/d_o = h_i/h_o

M = -(-30)/10 = 30/10 = 3 times

The image height (h_i) :

M = h_i / h_o

3 = h_i / 1 cm

h_i = 3 (1 cm)

h_i = 3 cm

the image height is 3 cm. The plus sign indicates that the image upward.

C. The properties of the image :

Virtual, upward, larger than object

5. The magnification of the image, according to the image below.

Known :

The object distance (d_o) = 60 cm

The focal length (f) = 20 cm

Wanted : The image magnification (M)

Solution :

The image distance :

1/f = 1/d_o + 1/d_i

1/d_i = 1/f – 1/d_o = 1/20 cm – 1/60 cm = 3/60 cm – 1/60 cm = 2/60 cm

d_i = 60/2 cm = 30 cm

The magnification of the image (M) :

M = d_i/d_o = 30 cm / 60 cm = 1/2 times

6. If the object is placed 6 cm from a concave mirror, the image distance is 12 cm as shown in figure below. Whhat is the image distance if the object is moved from the original position 1 cm away from the mirror.

Known :

The object distance (d_o) = 6 cm

The image distance (d_i) = 12 cm

Wanted : if the object distance (d_o) = 7 cm then the image distance is …

Solution :

1/f = 1/do + 1/di = 1/6 + 1/12 = 2/12 + 1/12 = 3/12

f = 12/3 = 4 cm

The focal length is positive, means that the focal point is real or the rays pass through the point.

The image distance :

1/d_i = 1/f – 1/d_o = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

d_i = 28/3 = 9.3 cm

7. A dentist observes and checks the patient’s teeth using a mirror with an 8 cm radius. In order for the hole to be seen clearly by the doctor, what is the distance between the patient’s teeth and the mirror?

A. less than 4 cm in front of a concave mirror

B. less than 4 cm in front of a convex mirror

C. more than 4 cm in front of the concave mirror

D. more than 4 cm in front of the convex mirror

Known :

Radius of mirror (r) = 8 cm

The focal length of mirror (f) = r / 2 = 8 / 2 = 4 cm

Wanted : The distance between the patient’s teeth and the mirror

Solution :

The mirror used is a concave mirror or a convex mirror? In order for the tooth hole to be clearly visible by the doctor, the mirror used should be able to enlarge the image of the tooth and the image must be upright. Convex mirror always produces inverted images and the size of the image is smaller than the size of the object. Conversely a concave mirror can produce an upright image if the object distance (d) is smaller than the focal length (f). If the object distance is greater than the focal length (f) then the concave mirror produces an inverted image.

The focal length (f) of the concave mirror is 4 cm, therefore the patient’s teeth should be less than 4 cm in front of a concave mirror.

The correct answer is A.

8. A concave mirror has a radius of curvature of 24 cm. If the object is placed 20 cm in front of the mirror then determine the properties of the image.

A. Real, upright and enlarged

B. Real, inverted and enlarged

C. Virtual, upright and enlarged

D. Virtual, inverted and smaller

Known :

Radius of curvature (r) = 24 cm

Focal length (f) = R/2 = 24/2 = +12 cm

The focal length of the concave mirror is positive or real because the light passes through the focal point of the mirror.

Object distance (d) = 20 cm

Wanted : Properties of image

Solution :

Image is virtual or real? Calculate the image distance (s’):

1/d + 1/d’ = 1/f

1/d’ = 1/f – 1/d

1/d’ = 1/12 – 1/20

1/d’ = 5/60 – 3/60

1/d’ = 2/60

d’ = 60/2

d’ = 30 cm

The image distance signed positive means that the image is real because it is passed by light.

Image enlarged ? Upright or inverted? First calculate the image magnification (M):

M = -d’ / d = -30/20 = -1.5

M > 1 means the image is enlarged, M has a negative sign means an inverted image. So the image properties are real, inverted, enlarged.

The correct answer is B.

9. A spherical mirror produces an image has size 5 times greater than the object on a screen, 5 meters away from the object. The mirror is…..

A. concave with the focal length of 25/24 m

B. convex with the focal length of 25/24 m

C. concave with the focal length of 24/25 m

D. convex with the focal length of 24/25 m

Known :

Magnification of image (M) = 5 times

The distance between object and image = 5 meters

Solution :

The size of the image produced by a convex mirror is always smaller than the size of the object, therefore, the mirror is a concave mirror.

Object distance (d) = x

Image distance (d’) = x + 5

Image magnification (M) = 5 times

The formula of image magnification :

The formula of the focal length (f) :

The correct answer is A.

[wpdm_package id=’858′]

1. Concave mirror problems and solutions
2. Convex mirror problems and solutions
3. Diverging lens problems and solutions
4. Converging lens problems and solutions
5. Optical instrument human eye problems and solutions
6. Optical instrument contact lenses problems and solutions
7. Optical instrument eyeglasses
8. Optical instrument magnifying glass problems and solutions
9. Optical instrument microscope – problems and solutions
10. Optical instrument telescopes problems and solutions

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1. 1-kg water at 100^oC mixed with 1-kg water at 10^oC in an isolated system. The specific heat of water is 4200 J/kg ^oC. Determine the final temperature of the mixture!

Known :

Mass of hot water (m₁) = 1 kg

Temperature of hot water (T₁) = 100^oC

Mass of cold water (m₂) = 1 kg

Temperature of cold water (T₂) = 10^oC

Wanted : The final temperature (T)

Solution :

Heat lost = Heat gained (isolated system)

m c ΔT = m c ΔT

m ΔT = m ΔT

m₁ (T₁ – T) = m₂ (T – T₂)

(1)(100 – T) = (1)(T – 10)

100 – T = T – 10

100 + 10 = T + T

110 = 2T

T = 110 / 2

T = 55

The final temperature is 55^oC.

2. A 3 kg block of lead at 80^o placed in 10 kg of water. The specific heat of lead is 1400 J.kg^-1C^-1 and the specific heat of water is 4200 J.kg^-1C^-1. The final temperature in thermal equilibrium is 20^oC. Determine the initial temperature of water!

Known :

Mass (m₁) = 3 kg

The specific heat of lead (c₁) = 1400 J.kg^-1C^-1

The temperature of lead (T₁) = 80 ^oC

Mass of water (m₂) = 10 kg

The specific heat of water (c₂) = 4200 J.kg^-1C^-1

The temperature of thermal equilibrium (T) = 20 ^oC

Wanted : The initial temperature of water (T₂)

Solution :

Heat lost = Heat gained

Q lead = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(3)(1400)(80-20) = (10)(4200)(20-T)

(4200)(60) = (42,000)(20-T)

252,000 = 840,000 – 42,000 T

42,000 T = 840,000 – 252,000

42,000 T = 588,000

T = 588,000 / 42,000

T = 14

The initial temperature of water is 14^oC.

3. A block of copper at 100^oC placed in 128 gram water at 30 ^oC. The specific heat of water is 1 cal.g^-1oC^-1 and the specific heat of copper is 0.1 cal.g^-1oC^-1. If the temperature of the thermal equilibrium is 36 ^oC, determine the mass of the copper!

Known :

The temperature of copper (T₁) = 100 ^oC

The specific heat of copper (c₁) = 0.1 cal.g^-1oC^-1

Mass of water (m₂) = 128 gram

Temperature of water (T₂) = 30 ^oC

The specific heat of water (c₂) = 1 cal.g^-1oC^-1

The temperature of thermal equilibrium (T) = 36 ^oC

Wanted : Mass of copper (m₁)

Solution :

Heat lost = Heat gained

Q copper = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(m₁)(0.1)(100-36) = (128)(1)(36-30)

(m₁)(0.1)(64) = (128)(1)(6)

(m₁)(6.4) = 768

m₁ = 768 / 6.4

m₁ = 120

The mass of copper is 120 gram.

4. A M-kg block of ice at 0^oC placed in 340-gram water at 20^oC in a vat. If the heat of fusion for water = 80 cal g^-1, the specific heat of water is 1 cal g^{-1 o}C^-1. All ice melts and the temperature of thermal equilibrium is 5^oC, determine the mass of ice!

Known :

Mass of water (m) = 340 gram

The temperature of ice (T _ice) = 0^oC

The temperature of water (T _water) = 20^oC

The temperature of thermal equilibrium (T) = 5^oC

The heat of fusion for water (L) = 80 cal g^-1

The specific heat of water (c _water) = 1 cal g^{-1 o}C^-1

Wanted : Mass of ice (M)

Solution :

Heat lost = Heat gained

Q water = Q ice

m c (ΔT) = m_es L_es + m c (ΔT)

(340)(1)(20-5) = M (80) + M (1)(5-0)

(340)(15) = 80M + 5M

5100 = 85M

M = 5100/85

M = 60 gram

[wpdm_package id=’714′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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4 Latent heat Heat of fusion Heat of vaporization – Problems and Solutions

1. Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 10³ J/kg.

Known :

Mass (m) = 1 gram = 1 x 10^-3 kg

Heat of fusion (L_F) = 64.5 x 10³ J/kg

Wanted : Heat (Q)

Solution :

Q = m L_F

Q = (1 x 10^-3 kg)(64.5 x 10³ J/kg)

Q = 64.5 Joule

2. Calculate the amount of heat released by 1 gram mercury to change phase from liquid to solid. Heat of fusion for mercury is 11.8 x 10³ J/kg.

Known :

Mass (m) = 1 gram = 1 x 10^-3 kg

Heat of fusion (L_F) = 11.8 x 10³ J/kg

Wanted : Heat (Q)

Solution :

Q = m L_F

Q = (1 x 10^-3 kg)(11.8 x 10³ J/kg)

Q = 11.8 Joule

3. Determine the amount of heat absorbed by 1 kg water to change phase from liquid to vapor (steam). Heat of vaporization for water = 2256 x 10³ J/kg

Known :

Mass (m) = 1 kg

Heat of vaporization (L_V) = 2256 x 10³ J/kg

Wanted : Heat (Q)

Solution :

Q = m L_V

Q = (1 kg)(2256 x 10³ J/kg)

Q = 2256 x 10³ Joule

4. Determine the amount of heat released by nitrogen to change phase from vapor to liquid. Heat of vaporization for nitrogen = 200 x 10³ J/kg

Known :

Mass (m) = 1 gram = 1 x 10^-3 kg

Heat of vaporization (L_V) = 200 x 10³ J/kg

Known : Heat (Q)

Solution :

Q = m L_V

Q = (1 x 10^-3 kg)(200 x 10³ J/kg)

Q = 200 Joule

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

Read more

1. A body with mass 2 kg absorbs heat 100 calories when its temperature raises from 20^oC to 70^oC. What is the specific heat of the body?

Known :

Mass (m) = 2 kg = 2000 gr

Heat (Q) = 100 cal

The change in temperature (ΔT) = 70^oC – 20^oC = 50^oC

Wanted : The specific heat (c)

Solution :

c = Q / m ΔT

c = 100 cal / (2000 gr)(50^oC)

c = 100 cal / 100,000 gr ^oC

c = 10² cal / 10⁵ gr ^oC

c = (10² cal)(10^-5 gr^{-1 o}C^-1)

c = 10^-3 cal gr^{-1 o}C^-1

c = 10^-3 cal/gr ^oC

The specific heat of the body is 10^-3 cal/gr ^oC

2. The specific heat of water is 4180 J/kg C^o. How much the heat capacity of 2 kg water…

Known :

The specific heat (c) = 4180 J/kg C^o

Mass (m) = 2 kg

Wanted : Heat capacity (C)

Solution :

C = m c

C = (2 kg)(4180 J/kg C^o)

C = (2)(4180 J/C^o)

C = 8360 J/C^o

3. The specific heat of aluminum is 900 J/kg C^o. How much the heat capacity of 2 gram aluminum…..

Known :

The specific heat of aluminum (c) = 900 J/kg C^o = 9 x 10² J/kg C^o

Mass (m) = 2 gram = 2/1000 kg = 2/10³ kg = 2 x 10^-3 kg

Wanted : Heat capacity (C)

Solution :

C = m c

C = (2 x 10^-3 kg)(9 x 10² J/kg C^o)

C = 18 x 10^-3 x 10² J/C^o

C = 18 x 10^-1 J/C^o

C = 1.8 J/C^o

[wpdm_package id=’710′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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1. 2 kcal (kilocalorie) = ….. calorie ?

Solution

1 kcal = 1000 calorie

2 kcal = 2 (1000 calorie) = 2000 calorie

2. 4 Calorie = ….. calorie ?

Solution

1 Calorie (a Capital C) = 1 kcal = 1000 calorie

4 Calorie = 4 (1000 calorie) = 4000 calorie

3. 10 calorie = ….. Joule ?

Solution

1 calorie = 4.186 Joule

10 calorie = 10 (4.186 Joule) = 41.86 Joule

4. 5 kcal = ….. Joule ?

Solution

1 kcal = 1000 calorie = 4186 Joule

5 kcal = 5 (4186 Joule) = 20930 Joule

5. 2000 Joule = ….. kcal ?

Solution

4186 Joule = 1 kcal

8372 Joule = 8372 / 4186 = 2 kcal

2000 Joule = 2000 / 4186 = 0.4777 kcal

[wpdm_package id=’706′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, the heat of fusion, the heat of vaporization
9. Energy conservation for heat transfer

Read more

9 Heat Mass Specific heat The change in temperature – Problems and Solutions

1. A 2 kg lead is heated from 50^oC to 100^oC. The specific heat of lead is 130 J.kg^{-1 o}C^-1. How much heat is absorbed by the lead?

Known :

Mass (m) = 2 kg

The specific heat (c) = 130 J.kg^-1C^-1

The change in temperature (ΔT) = 100^oC – 50^oC = 50^oC

Wanted : Heat (Q)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = the specific heat, ΔT = the change in temperature

The heat absorbed by lead :

Q = (2 kg)(130 J.kg^-1C^-1)(50^oC)

Q = (100)(130)

Q = 13,000 Joule

Q = 1.3 x 10⁴ Joule

2. The specific heat of copper is 390 J/kg ^oC, the change in temperature is 40^oC. If the copper absorbs 40 Joule of heat, what is the copper’s mass!

Known :

The specific heat of copper (c) = 390 J/kg^oC

The change in temperature (ΔT) = 40^oC

Heat (Q) = 40 J

Wanted : Mass (m) of copper

Solution :

Q = m c ΔT

40 J = (m)(390 J/kg ^oC)(40^oC)

40 = (m)(390 /kg)(40)

40 = (m)(390 /kg)(4)

40 = (m)(1560 /kg)

m = 40 / 1560

m = 0.026 kg

m = 26 gram

3. The initial temperature of 20 gram water is 30^oC. The specific heat of water is 1 cal g^{-1 o}C^-1. If water absorbs 300 calories of heat, determine the final temperature!

Known :

Mass (m) = 20 gr

The initial temperature (T₁) = 30^oC

The specific heat of water (c) = 1 cal gr^{-1 o}C^-1

Heat (Q) = 300 cal

Wanted : The final temperature of water

Solution :

Q = m c ΔT

300 cal = (20 gr)(1 cal gr^{-1 o}C^-1)(T₂-30)

300 = (20)(1)(T₂-30)

300 = 20 (T₂-30)

300 = 20T₂ – 600

300 + 600 = 20T₂

900 = 20T₂

T₂ = 900 / 20

T₂ = 45

The change in temperature is 45^oC – 30^oC = 15^oC.

4. The change in temperature of the sea water is 1^oC when water absorbs 3900 Joule of heat. The specific heat of the sea water is 3.9 × 10³ J/kg°C, what is the mass of the sea water.

Known :

The change in temperature (ΔT) = 1^oC

Heat (Q) = 3900 Joule

The specific heat of the sea water (c) = 3.9 x 10³ J/kg°C = 3900 J/Kg°C

Wanted : Mass (m)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature

m = Q / c ΔT = 3900 / (3900)(1) = 3900 / 3900 = 1 kg

5. A 2-kg copper absorbs 39,000 J of heat at 30°C. If the specific heat of copper is 390 J/kg °C, what is the final temperature of the copper…

Known :

Mass (m) = 2 kg

Initial temperature (T₁) = 30^oC

Heat (Q) = 39,000 Joule

Specific heat (c) of copper = 390 J/kg ^oC

Wanted : The final temperature (T₂)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature

Q = m c ΔT = m c (T₂ – T₁)

39,000 = (2)(390)(T₂ – 30)

100 = (2)(1)(T₂ – 30)

100 = (2)(T₂ – 30)

50 = T₂ – 30

T₂ = 50 + 30

T₂ = 80^oC

6. A 5-kg water is heated from 15°C to 40°C. What is the heat is absorbed by water. Specific heat of water is 4.2 × 10³ J/Kg° C.

Known :

Mass (m) = 5 kg

Initial temperature (T₁) = 15°C

Final temperature (T₂) = 40°C

Specific heat of water (c) = 4.2 × 10³ J/kg° C

Wanted : Heat (Q)

Solution :

Q = m c ΔT

Q = (5 kg)(4.2 × 10³ J/kg°C)(40°C – 15°C)

Q = (5)(4.2 × 10³ J)(25)

Q = 525 x 10³ J

Q = 525,000 Joule

7. A 2-kg water is heated from 24°C to 90°C. What is the heat is absorbed by water. Specific heat of water is 4.2 × 10³ J/Kg° C.

Known :

Mass (m) = 2 kg

Initial temperature (T₁) = 24°C

Final temperature (T₂) = 90°C

Specific heat of water (c) = 4,200 Joule/kg°C

Wanted :: Heat (Q)

Solution :

Q = m c ΔT

Q = (2 kg)(4,200 Joule/kg°C)(90°C – 24°C)

Q = (2 kg)(4,200 Joule/kg°C)(66°C)

Q = (132)(4,200 Joule)

Q = 554,400 Joule

8. A 5-gram water is heated from 10°C to 40°C. What is the heat is absorbed by water. Specific heat of water is 1 × 10³ cal/gr° C.

Known :

Mass (m) = 5 gram

Initial temperature (T₁) = 10^oC

Final temperature (T₂) = 40^oC

Specific heat of water (c) = 1 cal/ gr°C

Wanted : Heat

Solution :

Q = m c ΔT

Q = (5 gram)(1 cal/ gr°C)(40^oC – 10^oC)

Q = (5)(1 cal)(30)

Q = 150 calories

9. A 0.2-kg water absorbs 42,000 Joule of heat at 25^oC. The specific heat of water is 4200 J/kg ^oC, what is the final temperature of water.

Known :

Mass of water (m) = 0.2 kg

Heat (Q) = 42,000 Joule

Specific heat of water (c) = 4200 J/kg ^oC

Initial temperature (T₁) = 25^oC

Wanted : Final temperature (T₂)

Solution :

Q = m c ΔT = m c (T₂ – T₁)

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature, T₁ = the initial temperature, T₂ = the final temperature

Q = m c (T₂ – T₁)

42,000 = (0.2)(4200)(T₂ – 25)

42,000 = 840 (T₂ – 25)

42,000 = 840 T₂ – 21,000

42,000 + 21,000 = 840 T₂

63,000 = 840 T₂

T₂ = 63,000 / 840

T₂ = 75^oC

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

Read more

Linear expansion is experienced only by solid objects; volume expansion is experienced by all objects, both solid, liquid, and gas. The equation of volume expansion is similar to the equation of linear expansion.

Description: V_o = Initial volume, V = Final volume, ΔV = V – V_o = The change in volume, T_o = Initial temperature, T = Final temperature, ΔT = T – T_o = The change in temperature, β = the coefficient of volume expansion. Unit of β = (C^o) ^-1

The above volume expansion equation applies only when the changes in the volume of the objects (both solid, liquid, and gas) are smaller than the original volume of the object. If the change in volume of an object is greater than the initial volume of the object, the equation of the volume expansion does not give the right results. Usually, the changes in volume experienced by solid objects are not too large. Conversely, the coefficient of volume expansion of the liquid and gas is large. The coefficient of volume expansion for gaseous substances is also easy to change if the temperature changes. Therefore the formula above is used only for the expansion of solid objects.

1. At 30 ^oC the volume of an aluminum sphere is 30 cm³. The coefficient of linear expansion is 24 x 10^{-6 o}C^-1. If the final volume is 30.5 cm³, what is the final temperature of the aluminum sphere?

Known :

The coefficient of linear expansion (α) = 24 x 10^{-6 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 24 x 10^{-6 o}C^-1 = 72 x 10^{-6 o}C^-1

The initial temperature (T₁) = 30^oC

The initial volume (V₁) = 30 cm³

The final volume (V₂) = 30.5 cm³

The change in volume (ΔV) = 30.5 cm³ – 30 cm³ = 0.5 cm³

Wanted : The final temperature (T₂)

Solution :

ΔV = β (V₁)(ΔT)

ΔV = β (V₁)(T₂ – T₁)

0.5 cm³ = (72 x 10^{-6 o}C^-1)(30 cm³)(T₂ – 30^oC)

0.5 = (2160 x 10^-6)(T₂ – 30)

0.5 = (2.160 x 10^-3)(T2 – 30)

0.5 = (2.160 x 10^-3)(T₂ – 30)

0.5 / (2.160 x 10^-3) = T₂ – 30

0.23 x 10³ = T₂ – 30

0.23 x 1000 = T₂ – 30

230 = T₂ – 30

230 + 30 = T₂

T₂ = 260^oC

2. The coefficient of linear expansion of a metal sphere is 9 x 10^{-6 o}C^-1. The internal diameter of the metal sphere at 20 ^oC is 2.2 cm. If the final diameter is 2.8 cm, what is the final temperature!

Known :

The coefficient of linear expansion (α) = 9 x 10^{-6 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10^{-6 o}C^-1 = 27 x 10^{-6 o}C^-1

The initial temperature (T₁) = 20^oC

The initial diameter (D₁) = 2.2 cm

The final diameter (D₂) = 2.8 cm

The initial radius (r₁) = D₁ / 2 = 2.2 cm³ / 2 = 1.1 cm³

The final radius (r₂) = D₂ / 2 = 2.8 cm³ / 2 = 1.4 cm³

The initial volume (V₁) = 4/3 π r₁³ = (4/3)(3.14)(1.1 cm)³ = (4/3)(3.14)(1.331 cm³) = 5.57 cm³

The final volume (V₂) = 4/3 π r₂³ = (4/3)(3.14)(1.4 cm)³ = (4/3)(3.14)(2.744 cm³) = 11.48 cm³

The change in volume (ΔV) = 11.48 cm³ – 5.57 cm³ = 5.91 cm³

Wanted : The final temperature (T₂)

Solution :

ΔV = β (V₁)(ΔT)

5.91 cm³ = (27 x 10^{-6 o}C^-1)(5.57 cm³)(T₂ – 20^oC)

5.91 = (150.39 x 10^-6)(T₂ – 20)

5.91 / 150.39 x 10^-6 = T₂ – 20

0.039 x 10⁶ = T₂ – 20

39 x 10³ = T₂ – 20

39,000 = T₂ – 20

39,000 + 20 = T₂

T₂ = 39,020 ^oC

3. A 2000-cm³ aluminum container, filled with water at 0^oC. And then heated to 90^oC. If the coefficient of linear expansion for aluminum is 24 x 10^-6 (^oC)^-1 and the coefficient of volume expansion for water is 6.3 x 10^-4 (^oC)^-1, determine the volume of spilled water.

Known :

The initial volume of the aluminum container and water (V_o) = 2000 cm³ = 2 x 10³ cm³

The initial temperature of the aluminum container and water (T₁) = 0^oC

The final temperature of the aluminum container and water (T₂) = 90^oC

The coefficient of linear expansion for aluminum (α) = 24 x 10^-6 (^oC)^-1

The coefficient of volume expansion for aluminum (γ) = 3α = 3 (24 x 10^-6 (^oC)^-1 ) = 72 x 10^{-6 o}C^-1

The coefficient of volume expansion for water (γ) = 6.3 x 10^-4 (^oC)^-1

Wanted : The volume of spilled water

Solution :

The equation of the volume expansion :

V = V_o + γ V_o ΔT

V – V_o = γ V_o ΔT

ΔV = γ V_o ΔT

V = final volume, V_o = initial volume, ΔV = the change in volume, γ = the coefficient of volume expansion, ΔT = the change in temperature

Calculate the change in volume of the aluminum container :

ΔV = γ V_o ΔT = (72 x 10^-6)(2 x 10³)(90) = 12960 x 10^-3 = 12.960 cm³

Calculate the change in volume of the water :

ΔV = γ V_o ΔT = (6.3 x 10^-4)(2 x 10³)(90) = 1134 x 10^-1 = 113.4 cm³

The change in volume of the water is greater than the aluminum container so that some water spilled.

Calculate the volume of spilled water :

113.4 cm³ – 12.960 cm³ = 100.44 cm³

[wpdm_package id=’702′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, the heat of fusion, the heat of vaporization
9. Energy conservation for heat transfer

Read more

1. At 20 ^oC, the length of a sheet of steel is 50 cm and the width is 30 cm. If the coefficient of linear expansion for steel is 10^{-5 o}C^-1, determine the change in area and the final area at 60 ^oC.

Known :

The initial temperature (T₁) = 20^oC

The final temperature (T₂) = 60^oC

The change in temperature (ΔT) = 60^oC – 20^oC = 40^oC

The initial area (A₁) = length x width = 50 cm x 30 cm = 1500 cm²

The coefficient of linear expansion for steel (α) = 10^{-5 o}C^-1

The coefficient of area expansion for steel (β) = 2α = 2 x 10^{-5 o}C^-1

Wanted : The change in area (ΔA)

Solution :

The change in area (ΔA) :

ΔA = β A₁ ΔT

ΔA = (2 x 10^{-5 o}C^-1)(1500 cm²)(40^oC)

ΔA = (80 x 10^-5)(1500 cm²)

ΔA = 120,000 x 10^-5 cm²

ΔA = 1.2 x 10⁵ x 10^-5 cm²

ΔA = 1.2 cm²

The final area (A₂) :

A₂ = A₁ + ΔA

A₂ = 1500 cm² + 1.2 cm²

A₂ = 1501.2 cm²

2. At 30 ^oC, the area of a sheet of aluminum is 40 cm² and the coefficient of linear expansion is 24 x 10^-6 /^oC. Determine the final temperature if the final area is 40.2 cm².

Known :

The initial temperature (T₁) = 30^oC

The coefficient of linear expansion (α) = 24 x 10^{-6 o}C^-1

The coefficient of area expansion (β) = 2a = 2 x 24 x 10^{-6 o}C^-1 = 48 x 10^{-6 o}C^-1

The initial area (A₁) = 40 cm²

The final area (A₂) = 40.2 cm²

The change in area (ΔA) = 40.2 cm² – 40 cm² = 0.2 cm²

Wanted : Determine the final temperature (T₂)

Solution :

Formula of the change in area (ΔA) :

ΔA = β A₁ ΔT

The final temperature (T₂) :

ΔA = β A₁ (T₂ – T₁)

0.2 cm² = (48 x 10^{-6 o}C^-1)(40 cm²)(T₂ – 30^oC)

0.2 = (1920 x 10^-6)(T₂ – 30)

0.2 = (1.920 x 10^-3)(T₂ – 30)

0.2 = (2 x 10^-3)(T₂ – 30)

0.2 / (2 x 10^-3) = T₂ – 30

0.1 x 10³ = T₂ – 30

1 x 10² = T₂ – 30

100 = T₂ – 30

100 + 30 = T₂

T₂ = 130

The final temperature = 130^oC

3. The radius of a ring at 20 ^oC is 20 cm. If the final radius at 100 ^oC is 20.5 cm, determine the coefficient of area expansion and the coefficient of linear expansion…

Known :

The initial temperature (T₁) = 30^oC

The final temperature (T₂) = 100^oC

The change in temperature (ΔT) = 100^oC – 30^oC = 70^oC

The initial radius (r₁) = 20 cm

The final radius (r₂) = 20.5 cm

Wanted : The coefficient of area expansion (β)

Solution :

The initial area (A₁) = π r₁² = (3.14)(20 cm)² = (3.14)(400 cm²) = 1256 cm²

The final area (A₂) = π r₂² = (3.14)(20.5 cm)² = (3.14)(420.25 cm²) = 1319.585 cm²

The change in area (ΔA) = 1319.585 cm² – 1256 cm² = 63.585 cm²

Formula of the change in area (ΔA) :

ΔA = β A₁ ΔT

The coefficient of area expansion :

ΔA = β A₁ ΔT

63.585 cm² = b (1256 cm²)(70 ^oC)

63.585 = b (87,920 ^oC)

β = 63.585 / 87,920 ^oC

β = 0.00072 /^oC

β = 7.2 x 10^-4 /^oC

β = 7.2 x 10^{-4 o}C^-1

The coefficient of linear expansion (α) :

β = 2 α

α = β / 2

α = (7.2 x 10^-4) / 2

α = 3.6 x 10^{-4 o}C^-1

[wpdm_package id=’698′]

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

Read more