100 g of water at 373 K are mixed with 055kg of ice at 253 K

100 g of water at 373 K are mixed with .055kg of ice at 253 K. What is the equilibrium temperature?

Solution

Here ,

let the equilibrium temperature is Tf

specific heat of ice . Sice = 2.108 J/gm*K

heat lost by water = heat gain by ice

55 * (334 + 2.108 * (273- 253) + (Tf - 273) * 4.186) = 100 * 4.186 * (373 - Tf)

solving for Tf

Tf = 305.6 K

the equilibrium temperature is 305.6 K