1. If a 2kW heater is immersed in 1kg of water at 12°C and is switched on for 10 minutes, find the amount of water boiled away.
2. A copper can of mass 0.1 kg contains 0.25kg of water at 60°C. If 0.1kg of ice at 0°C is added, find the resulting temperature. (Specific heat capacity of copper = 400Jkg^-1 K^-1)
3. 0.1kg of a mixture of ice and water is placed in a polystyrene cup. When 0.15kg of water at 40°C is added, the resulting temperature is 10°C. Find the mass of ice in the mixture.
4. 0.3kg of ice is dropped into 0.1kg of water at 30°C. As a result, only part of the ice is melted.
a)What is the final temperature of the mixture?
b) Find the mass of ice melted.
Physics (Heat)
2009-12-30 3:52 am
回答 (3)
2009-12-31 7:17 am
✔ 最佳答案
1. By E = Pt
The enegy provided by the heater
= 2 x 1000 x ( 60 x 10) = 1200000J
By E = mcT
The energy required to boil water from 12 C to 100 C
= (1)(4200)(100 - 12) +
= 369600 J
Let the mass of water boiled away be m kg
E = lm
m (2.26 x 10^6) = 1200000 - 369600
m = 0.367 (3 sig. fig)
Therefore, 0.367 kg of water is boiled away
2.
Let T degC be the final temperature
By the conservation of energy,
Energy gained = Energy loss
(0.1) (3.34 x 10^5) + (0.1)(4200)(T - 0)=(0.1)(400)(60 - T)+(0.25)(4200)(60 - T)
33400 = 2400 - 40T + 63000 - 1050T
1090 T = 32000
T = 29.4 (3 sig.fig)
Therefore, the final temperature is 29.4 degC
3.
Let the mass of ice be m kg
By the conservation of energy,
Energy gained = Energy loss
m (3.34 x 10^5) + m (4200) (10 - 0) + (0.1 - m) (4200) (10 - 0)
= (0.15)(4200)(40 - 10)
334000m + 42000m + 4200 - 42000m = 18900
334000 m = 14700
m = 0.0440 (3 sig. fig)
Therefore, the mass of ice is 0.044kg
4.
(a) 0 degC
(b)
Let the mass of ice melted be m kg
By the conservation of energy,
Energy gained = Energy loss
m (3.34 x 10^5) = (0.1) (4200) (30 - 10)
m = 0.0251
Therefore, 0.0251 kg of ice melt
2009-12-31 6:23 pm
2009-12-30 4:26 am
1. Heat output by heater = 2000 x 10 x 60 J = 1 200 000 J
Mass of water boiled away = 1 200 000/(2260000 + 4200 x(100-12)) kg
2. Let T be the final temprature
0.1 x 400 x(60-T) + 0.25 x 4200 x (60-T) = 0.1 x (334000 + 4200 x T)
solve for T
3. Heat given out by water at 40'C
= 0.15 x 4200 x (40-10) J
Heat absorbed by ice = m.[334000 + 4200 x 10] J, where m is the mass of ice in the mixture
Heat absorbed by water at 0'C = (0.1 - m) x 4200 x 10 J
Hence, 0.15 x 4200 x (40-10) = m.[334000 + 4200 x 10] + (0.1 - m) x 4200 x 10
solve for m
4. (a) 0'C
(b) Heat given out by water = 0.1 x 4200 x 30 J
Mass of ice melted = ( 0.1 x 4200 x 30 )/334000 kg = 0.04 kg
Mass of water boiled away = 1 200 000/(2260000 + 4200 x(100-12)) kg
2. Let T be the final temprature
0.1 x 400 x(60-T) + 0.25 x 4200 x (60-T) = 0.1 x (334000 + 4200 x T)
solve for T
3. Heat given out by water at 40'C
= 0.15 x 4200 x (40-10) J
Heat absorbed by ice = m.[334000 + 4200 x 10] J, where m is the mass of ice in the mixture
Heat absorbed by water at 0'C = (0.1 - m) x 4200 x 10 J
Hence, 0.15 x 4200 x (40-10) = m.[334000 + 4200 x 10] + (0.1 - m) x 4200 x 10
solve for m
4. (a) 0'C
(b) Heat given out by water = 0.1 x 4200 x 30 J
Mass of ice melted = ( 0.1 x 4200 x 30 )/334000 kg = 0.04 kg
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