Kutentha Kwambiri Kutentha Kwapadera Kusintha kwa kutentha - Mavuto ndi Mayankho

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

1. A 2 kg lead is heated from 50^oC kuti 100^oC. The kutentha kwapadera of lead is 130 J.kg^{-1 o}C^-1. Zingati kutentha is absorbed by the lead?

Zodziwika:

Misa (m) = 2 kg

Kutentha kwenikweni (c) = 130 J.kg^-1C^-1

Kusintha kwa kutentha (ΔT) = 100^oC - 50^oC=50^oC

Akufuna: kutentha (Q)

Yankho:

Q = mc ΔT

Q= kutentha, m = mass, c = the specific heat, ΔT= kusintha kwa kutentha

The heat absorbed by lead :

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

Q = (100)(130)

Q = 13,000 ma joule

Q = 1.3 X XUMUM⁴ 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!

Zodziwika:

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

Kusintha kwa kutentha (ΔT) = 40^oC

Heat (Q) = 40 y

Akufuna: Misa (m) cha mkuwa

Yankho:

Q = mc ΔT

40 y = (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 makilogalamu

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!

Zodziwika:

Mass (m) = 20 gr

Kutentha koyambirira (T₁= = 30^oC

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

kutentha (Q) = 300 cal

Akufuna: The final temperature of water

Yankho:

Q = mc .T

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.

Zodziwika:

Kusintha kwa kutentha (ΔT) = 1^oC

kutentha (Q) = 3900 Joule

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

Akufuna: Misa (m)

Yankho:

Q = mc ΔT

Q= kutentha, m = mass, c = kutentha kwapadera, ΔT= kusintha kwa kutentha

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…

Zodziwika:

Kulemera (m) = 2 kg

Kutentha koyambirira (T₁= = 30^oC

kutentha (Q) = 39,000 Joule

Kutentha kwapadera (c) cha mkuwa = 390 J/kg ^oC

anafuna : Kutentha komaliza (T₂)

Yankho:

Q = mc .T

Q= kutentha, m = mass, c = kutentha kwapadera, ΔT = kusintha kwa kutentha

Q = m c ΔT = m c (T₂ - T₁)

39,000 = (2)(390)(T₂ - 30)

100 = (2)(1)(T)₂ - 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.

Zodziwika:

Kulemera (m) = 5 kg

Kutentha koyambirira (T₁) = 15°C

Final temperature (T₂) = 40°C

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

Akufuna: kutentha (Q)

Yankho:

Q = mc .T

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 Joules

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.

Zodziwika:

Kulemera (m) = 2 kg

Kutentha koyambirira (T₁) = 24°C

Final temperature (T₂) = 90°C

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

Wanted :: kutentha (Q)

Yankho:

Q = m c .T

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 Joules

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 X 10³ kal/gr° C.

Zodziwika:

Kulemera (m) = magalamu 5

Kutentha koyambirira (T₁= = 10^oC

Final temperature (T₂= = 40^oC

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

anafuna : Kutentha

Yankho:

Q = mc .T

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

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

Q = 150 makilogalamu

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.

Zodziwika:

Misa zamadzi (m) = 0.2 kg

kutentha (Q) = 42,000 Joule

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

Kutentha koyambirira (T₁= = 25^oC

Akufuna: Final temperature (T₂)

Yankho:

Q = mc ΔT = m c (T₂ - T₁)

Q= kutentha, m = mass, c = kutentha kwapadera, ΔT = kusintha kwa kutentha, T₁ = the initial temperature, T₂ = the final temperature

Q = mc (T₂ - T₁)

42,000 = (0.2)(4200)(T)₂ - 25)

42,000 = 840 (T)₂ - 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. Kusintha miyeso ya kutentha
2. Kukula kwa mzere
3. Kukula kwa dera
4. Kukula kwa voliyumu
5. kutentha
6. Kutentha kofanana ndi makina
7. Kutentha ndi mphamvu ya kutentha yeniyeni
8. Kutentha kobisika, kutentha kwa fusion, kutentha kwa nthunzi
9. Kusunga mphamvu potumiza kutentha