10 kg of ice at -10 C is added to 100 kg of water to lower its temperature from 25 C. Consider no heat exchange to surroundings. The decrement to the temperature of water is C. (specific heat of ice = 2100 J/kg C, specific heat of water = 4200 J/kg C, latent heat of fusion of ice = 3.36 105 J/kg)
- 15
- 10
- 6.67
- 11.6
Answer
(2) 10
Solution
Heat to warm the ice from -10 C to 0 C:
Q1 = 10 2100 10 = 2.1 105 J
Heat to melt the ice: Q2 = 10 3.36 105 = 3.36 106 J
Q1 + Q2 = 3.57 106 J
Heat the water can give in cooling to 0 C: 100 4200 25 = 1.05 107 J (larger)
All the ice melts. Let the final temperature be T.
Heat lost by water = Heat gained by ice
100 4200 (25 - T) = 3.57 106 + 10 4200 T
1.05 107 - 4.2 105 T = 3.57 106 + 4.2 104 T
4.62 105 T = 6.93 106
T = 15 C
Decrease in temperature of water = 25 - 15 = 10 C
Answer: (2)