Question

A 0.0575 kg ice cube at −30.0°C is placed in 0.617 kg of 35.0°C water in a very well insulated container, like the kind we used in class. The heat of fusion of water is 3.33 x 105 J/kg, the specific heat of ice is 2090 J/(kg · K), and the specific heat of water is 4190 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?

Answer 1

Here ,

let the final temperature of the system is T

heat lost by water = heat gain by ice

0.0575 * (2090 * 30 + 3.35 *10^5 + 4190 * T) = 0.617 * 4190 * (35 - T)

solving for T

T = 23.9 degree C

the final temperature of the system is 23.9 degree C