9 Heat Mass Specific heat The change in temperature – Problems and Solutions
1. A 2 kg lead is heated from 50o100에 CoC. 비열 of lead is 130 J.kg영하 1도C-1. 얼마 is absorbed by the lead?
알려진 바에 따르면:
질량 (m) = 2kg
The specific heat (c) = 130 J.kg-1C-1
온도의 변화 (ΔT) = 100oC - 50oC = 50oC
구함: 열 (큐)
솔루션 :
Q = mc ΔT
질문 = , m = mass, c = the specific heat, ΔT = 온도의 변화
The heat absorbed by lead :
Q = (2 kg)(130 J.kg-1C-1)(삼oC)
Q = (100)(130)
질문 = 13,000 줄 (Joule)
Q = 1.3 X 104 줄
2. The specific heat of copper is 390 J/kg oC, the change in temperature is 40oC. If the copper absorbs 40 Joule of heat, what is the copper’s mass!
알려진 바에 따르면:
The specific heat of copper (c) = 390 J/kgoC
온도의 변화 (ΔT) = 40oC
Heat (Q) = 40 J
구함: 질량(m) 구리
솔루션 :
Q = mc ΔT
40 J = (중)(390 J/kg oC)(40oC)
40 = (m)(390 /kg)(40)
40 = (m)(390 /kg) (4)
40 = (m)(1560 /kg)
엠 = 40 / 1560
엠 = 0.026의 kg
m = 26 gram
3. The initial temperature of 20 gram water is 30oC. The specific heat of water is 1 cal g-1 oC-1. If water absorbs 300 calories of heat, determine the final temperature!
알려진 바에 따르면:
Mass (m) = 20 gr
초기 온도 (T1) = 30oC
The specific heat of water (c) = 1 cal gr-1 oC-1
열 (Q) = 300 cal
구함: The final temperature of water
솔루션 :
Q = mc ΔT
300 cal = (20 gr)(1 cal gr-1 oC-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 45oC - 30oC = 15oC.
4. The change in temperature of the sea water is 1oC 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.
알려진 바에 따르면:
온도의 변화 (ΔT) = 1oC
열 (Q) = 3900 Joule
The specific heat of the sea water (c) = 3.9 x 103 J/kg°C = 3900 J/Kg°C
구함: 질량 (m)
솔루션 :
Q = mc ΔT
질문 = , m = mass, c = 비열, ΔT = 온도의 변화
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…
알려진 바에 따르면:
질량(m) = 2 kg
초기 온도 (T1) = 30oC
열 (Q) = 39,000 Joule
비열 (C) 구리 = 390 J/kg oC
구함 : 최종 온도 (T2)
솔루션 :
Q = mc ΔT
질문 = , m = mass, c = 비열, ΔT = 온도의 변화
Q = mc ΔT = mc (T2 - T1)
39,000 = (2)(390)(T2 - 30)
100 = (2)(1)(T2 - 30)
100 = (2)(T2 - 30)
50 = 티2 - 30
T2 = 50 + 30
T2 = 80oC
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.
알려진 바에 따르면:
질량(m) = 5 kg
초기 온도 (T1) = 15°C
최종 온도 (T2) = 40°C
Specific heat of water (c) = 4.2 × 103 J/kg°C
구함: 열 (큐)
솔루션 :
Q = mc ΔT
Q = (5 kg)(4.2 × 103 J/kg°C)(40°C – 15°C)
질문 = (5)(4.2 × 103 J)(25)
Q = 525 x 103 J
Q = 525,000 줄
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.
알려진 바에 따르면:
질량(m) = 2 kg
초기 온도 (T1) = 24°C
최종 온도 (T2) = 90°C
Specific heat of water (c) = 4,200 Joule/kg°C
Wanted :: 열 (큐)
솔루션 :
Q = m c ΔT
Q = (2 kg)(4,200 줄/kg°C)(90°C – 24°C)
Q = (2 kg)(4,200 줄/kg°C)(66°C)
Q = (132)(4,200줄)
Q = 554,400 줄
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 라임/gr° C.
알려진 바에 따르면:
질량(m) = 5그램
초기 온도 (T1) = 10oC
최종 온도 (T2) = 40oC
Specific heat of water (c) = 1 cal/ gr°C
구함 : 열
솔루션 :
Q = mc ΔT
Q = (5그램)(1 cal/ gr°C)(40oC - 10oC)
질문 = (5)(1 cal)(30)
질문 = 150 칼로리
9. A 0.2-kg water absorbs 42,000 Joule of heat at 25oC. The specific heat of water is 4200 J/kg oC, what is the final temperature of water.
알려진 바에 따르면:
질량 물의 (m) = 0.2kg
열 (Q) = 42,000 Joule
Specific heat of water (c) = 4200 J/kg oC
초기 온도 (T1) = 25oC
구함: 최종 온도 (T2)
솔루션 :
Q = mc ΔT = mc (T2 - T1)
질문 = , m = mass, c = 비열, ΔT = 온도의 변화, T1 = the initial temperature, T2 = the final temperature
Q = mc (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 = 75oC
