Question

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

Answer 1

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;

T = 65.6999 degree celcius.