$5$kg of ice at $-10^o$C is added to $5$kg of water at $10^o$C. The temperature of the resulting mixture will.
- $-10^o$C
- $0^o$C
- $3.3^o$C
- $10^o$C
The heat lost by the water cooling to 0C is 5kg * 4186 J/kgK * 10K = 209,300 J. The heat required to warm ice to 0C is 5kg * 2100 J/kgK * 10K = 105,000 J. Since there is excess heat, the ice will melt partially, keeping the mixture at 0C.
To determine the final temperature, calculate the heat required to bring the 5 kg of water down to 0 degrees C, which is 5 kg multiplied by 1 kcal/kgC multiplied by 10C for a total of 50 kcal. The heat available from the water is more than enough to warm the 5 kg of ice to 0 degrees C, which requires 25 kcal (5 kg multiplied by 0.5 kcal/kgC multiplied by 10C), but not enough to melt all the ice, as melting would require 400 kcal. Because the heat from the water is used up before all the ice melts, the system reaches thermal equilibrium at the melting point of ice, which is 0 degrees C.