טעמפּעראַטור און היץ – פּראָבלעמען און לייזונגען
1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. קסנומקסOX = ….. oC.
באַקאַנט:
The freezing point of water = -30o
The boiling point of water = 90o
געוואָלט: 60oX = ….. oC
לייזונג:
On the Fahrenheit scale, the freezing point of water is 32oF and the boiling point of water is 212oF. Between the freezing point and the boiling point, 212o - קסנומקסo = קסנומקסo.
On the Celsius scale, the freezing point of water is 0oC and the boiling point of water is 100oC. Between the freezing point and the boiling point, 100o - קסנומקסo = קסנומקסo.
On the X scale, the freezing point of water is -30oX and the boiling point of water is 90oX. Between the freezing point and the boiling point, 90o – (-30o) = 90o + קסנומקסo = קסנומקסo.
Change the X scale to the Celsius scale :
2. A metal rod heated from 30oC צו 80oC. The final length of the rod is קסנומקס סענטימעטער. דער קאָואַפישאַנט פון לינעאַרע עקספּאַנסיע is 3.10-3 oC-1. What is the initial length of the metal rod?
באַקאַנט:
די ערשטע טעמפּעראַטור (T1) = 30oC
די לעצטע טעמפּעראַטור (T2) = 80oC
די ענדערונג אין טעמפּעראַטור (ΔT) = 80oC - 30oC=50oC
דער קאָעפיציענט פון לינעאַרער יקספּאַנשאַן (α) = 3.10-3 oC-1
The final length of the metal (L) = 115 cm
געוואָלט: The initial length of the metal rod (Lo)
לייזונג:
The equation of the linear expansion :
ל = לo + ΔL
ל = לo + α Lo ΔT
ל = לo (1 + α ΔT)
115 = לo (1 + 3.10-3.קסנומקס)
115 = לo (1 + 150.10-3)
115 = לo (1 + 0.15)
115 = לo (1.15)
Lo = 115/1.15
Lo = 100 סענטימעטער
3. The initial length of a brass rod is קסנומקס סענטימעטער. After heated, the final length of the brass is 40.04 cm and the final temperature is 80oC. If the coefficient of linear expansion of the brass is קסנומקס רענטגענ קסנומקס-5 oC-1, what is the initial temperature of the brass rod.
באַקאַנט:
די לעצטע טעמפּעראַטור (T2) = 80oC
די ערשטע לענג (Lo) = 40קם
The final length (L) = 40.04 cm
The increase in length (ΔL) = 40.04 cm – 40 cm = 0.04 cm
דער קאָעפיציענט פון לינעאַרער יקספּאַנשאַן (α) = 2.0 x 10-5 oC-1
לייזונג: ערשט טעמפּעראַטור (T1)
Solution ;
The equation of the linear expansion :
ל = לo + α Lo ΔT
L – Lo = α Lo ΔT
ΔL = α Lo ΔT
ΔL = α Lo (T2 - ה1)
0.04 = (2.0 קס 10-5)(40)(80 – T1)
0..04 = (80 x 10-5)(80 – ט1)
0.04 = 0.0008 (80 – T1)
0.04 = 0.064 – 0.0008 T1
0..0008 T1 = 0.064 - 0.040
קסנומקס ה1 = קסנומקס
T1 = קסנומקסoC
4. Two metal rods with the same size but different type, as shown in figure below. The thermal conductivity of metal I = 4 times the thermal conductivity of metal II. What is the temperature between both metals.
באַקאַנט:
The size of both rods is the same.
The thermal conductivity of metal I = 4k
The thermal conductivity of metal II = k
The temperature of the one end of metal I = 500 C
The temperature of the one end of metal II = 00 C
געוואלט: The temperature between both the metal rods
לייזונג:
The equation of the heat conduction :
Q/t = the rate of heat conduction, k = טערמאַל קאַנדאַקטיוואַטי, א = the cross-sectional area, ה1-T2 = די ענדערונג אין טעמפּעראַטור, l = the length of the rod.
The temperature between both rods :
The temperature at the center between both the metals rods is 40oC.
5.
(1) Conductivity of metal
(2) The difference of temperature
(3) The length of metal
(4) Mass of metal
Factors that determine the rate of heat conduction on metals are…..
לייזונג:
Based on the equation of heat conduction, factors that determine the rate of heat conduction on metals are conductivity of metal (k), the difference of temperature (T) and the length of metal (l).
6. Two rods of the same size but different type, as shown in the figure below. The thermal conductivity of rod P is 2 times the thermal conductivity of rod Q. What is the temperature between both rods.
באַקאַנט:
Both rods have the same size.
Thermal conductivity of rod P (kP) = 2k
The thermal conductivity of rod Q (kQ) = k
געוואלט: The temperature between both rods
לייזונג:
The equation of the heat conduction :
Q/t = the rate of heat conduction, k = טערמאַל קאַנדאַקטיוואַטי, א = the cross-sectional area, ה1-T2 = די ענדערונג אין טעמפּעראַטור, l = the length of the rod.
The temperature at the center between both rods :
8. 100-gram oil at 20oC and 50-gram iron at 75 oC are placed in 200-gram iron container. The increase in temperature of the container is 5oC and the specific heat of oil is 0.43 cal/g oC. What is the specific heat of the iron?
באַקאַנט:
Mass of iron container (m) = 200 gr
The initial temperature of the iron container (T1) = the temperature of oil = 20oC
The final temperature of the iron container (T2) = 20oC+5oC=25oC
Mass of oil (m) = 100 gram
The specific heat of oil (cייל) = 0.43 cal/g oC
The initial temperature of oil (T1) = 20oC
The final temperature of oil (T2) = 20oC+5oC=25oC
Mass of iron (m) = 50 gram
The initial temperature of oil (T1) = 75oC
The final temperature of oil (T2) = 25oC
געוואָלט: The specific heat of iron (c iron)
לייזונג:
Heat released by iron :
Q = m c ΔT = (50)(c)(75-25) = (50)(c)(50) = 2500c calorie
Heat absorbed by the iron container :
Q = m c ΔT = (200)(c)(25-20) = (200)(c)(5) = 1000c calorie
Heat absorbed by oil :
Q = m c ΔT = (100)(0.43)(25-20) = (43)(5) = 215 calorie
Black principle states that in a isolated system, heat released by the hotter object, absorbed by the cooler object.
Q release = Q absorb
2500c = 1000c + 215
2500c – 1000c = 215
1500c = 215
c = 215/1500
c = 0.143 cal/g oC
9. A 200-gram water at 20°C placed in 50-gram ice at -2°C. If the change of heat just between water and ice, what is the final temperature of the mixture? The specific heat of water is 1 cal/gr°C, the specific heat of ice is 0.5 cal/gr°C, the heat of fusion for ice is 80 cal/gr.
באַקאַנט:
Mass of water (mוואַסער) = 200 גראַם
Temperature of water (Tוואַסער) = 20oC
The specific heat of water (cוואַסער) = 1 cal/gr°C
Mass of ice (mייַז) = 50 גראַם
טעמפּעראַטור פון אייז (טייַז) = -2oC
The specific heat of ice (cייַז) = 0.5 cal/gr°C
The heat of fusion for ice (L) = 80 cal/gr
לייזונג:
Heat to increases ice from -2oC צו קסנומקסoC:
ק = מק Δט
Q = (50 gram)(0.5 cal/gr°C)(0oC – (-2oג))
Q = (50)(0.5 cal)(2)
Q = 50 calorie
Heat for melting all ice :
Q = m L = (50 gram)(80 cal/gram) = 4000 calorie
Heat for decrease temperature of all water from 20oC צו קסנומקסoC:
ק = מק Δט
Q = (200 gram)(1 cal/gr°C)(0oC – (20oג))
Q = (200)(1 cal)(-20)
Q = -4000 calorie
Plus sign indicates that the heat is added, the minus sign indicates that heat released.
50-calorie of heat needed to increase the temperature of ice to 0oC and 4000-calorie needed to melting all ice. Total heat = 4050 calorie. The heat released by water is 4000 calorie.
Some of the ice not melting, so the final temperature of ice and water is 0oC.
10. A 200-gram aluminum at 20oC placed in 100-gram water at 80oC in a container. The specific heat of aluminum is 0.22 cal/g oC and the specific heat of water is 1 cal/g oC. What is the final temperature of aluminum?
באַקאַנט:
Mass of aluminum = 200 gram
Temperature of aluminum = 20oC
Mass of water = 100 gram
Temperature of water = 80oC
The specific heat of aluminum = 0.22 cal/g oC
The specific heat of water = 1 cal/g oC
געוואלט: The final temperature of aluminum
לייזונג:
Aluminum and water in thermal equilibrium so that the final temperature of aluminum = the final temperature of water.
Heat released by hot water (Q release) = heat absorbed by aluminum (Q absorb)
mוואַסער ג (ΔT) = mאַלומינום ג (ΔT)
(100)(1)(80 – T) = (200)(0.22)(T – 20)
(100)(80 – T) = (44)(T – 20)
8000 – 100T = 44T – 880
8000 + 880 = 44T + 100T
8880 = 144T
ה = 62oC
11. A 50-gram metal at 85 °C placed in 50 gram water at 29.8 °C. The specific heat of water = 1 cal.g -1 .°C-1. The final temperature is 37 °C. What is the specific heat of metal.
באַקאַנט:
Mass of metal (mמעטאַל) = 50 גראַם
Temperature of metal = 85oC
Mass of water (mוואַסער) = 50 גראַם
Temperature of water = 29,8oC
The specific heat of water (cוואַסער) = 1 cal.g -1 .°C-1
The final temperature of water = 37oC
געוואָלט: The specific heat of metal (c metal)
לייזונג:
Heat released by hot metal (Q release) = heat absorbed by water (Q absorb)
mמעטאַל ג (ΔT) = mוואַסער ג (ΔT)
(50)(c)(85 – 37) = (50)(1)(37 – 29.8)
(c)(85 – 37) = (1)(37 – 29.8)
48 c = 7.2
c = 0.15 cal.g -1 .°C-1
12. A block of ice with mass of 50-gram at 0°C and 200-gram water at 30°C, placed in a container. . If the specific heat of water is 1 cal.g- קסנומקס ° C -קסנומקס and the heat of fusion for ice is 80 cal.g -1. What is the final temperature of the mixture.
באַקאַנט:
Mass of ice (mייַז) = 50 גראַם
The temperature of ice = 0°C
Mass of water (mוואַסער) = 200 גראַם
Temperature of water = 30oC
The specific heat of water (cוואַסער) = 1 cal.g- קסנומקס ° C -קסנומקס
The heat of fusion for ice (Lייַז) = 80 cal.g -קסנומקס
געוואלט: די לעצטע טעמפּעראַטור
לייזונג:
Estimate the final condition :
Heat released by water to decrease its temperature from 30oC צו קסנומקסoC:
Qמעלדונג = עםוואַסער cוואַסער (ΔT) = (200)(1)(30-0) = (200)(30) = 6000
Heat needed to melting all ice :
ק = מייַז Lייַז= (50)(80) = 4000
Heat used to melting all ice is 4000, while heat released by water is 6000. Can be concluded that the final temperature of the mixture above 0oC.
Black principle :
Heat released by water = heat for melting all ice + heat to increase the temperature of ice.
(mוואַסער)(cוואַסער)(ΔT) = (mייַז)(Lייַז) + (mייַז)(cוואַסער)(ΔT)
(200)(1)(30-T) = (50)(80) + (50)(1)(T-0)
(200)(30-T) = (50)(80) + (50)(T-0)
6000 – 200T = 4000 + 50T – 0
6000 – 4000 = 50T + 200T
2000 = 250T
T = 2000/250
ה = 8oC
