A 112g cube of ice at 0°C is added to 1kg of water at 82°C. The final temperature is calculated using conservation of energy and accounting for the latent heat of fusion of ice and specific heat of water. The energy lost by the water heating the ice equals the energy gained by melting and heating the ice. Solving the equation yields a final temperature of 65.6999°C.

1. A 112-g cube of ice at 0°C is dropped into 1.0 kg of water that was originally at 82°C. What
is the final temperature of the water after the ice has melted?
in C
Solution
Latent heat of fusion of ice L = 334 KJ/kg, specific heat of water s = 4.1813 KJ/Kg-k
conservation of energy;
Energy loss by the water = energy gained by the ice ;
Energy lost by the water at 82°C when the temperature drops to T°C
Elost= 1 kg x 4.1813 KJ/(kg.K) x (82 - T)°C ............................................(1)
The energy required to melt the ice at 0°C to water at 0°C
= 0.112 Kg x 334 KJ/kg (since Q = m*L)
= 37.408 KJ
The energy needed by the melted ice water to reach T°C from 0°C
= 0.112 kg x 4.1813 KJ/(kg.K) x T °C (since, Q = m*s*dt)
= (0.468 x T) kJ
therefore, total energy gain by the ice Egain = 37.408 KJ + (0.468 x T) KJ ...............................(2)
now applying conservation of energy;
Elost = Egain;
1 kg x 4.1813 kJ/(kg.K) x (82 - T)°C = 37.408 KJ + (0.468 x T) KJ;

2. T = 65.6999 degree celcius.