Question

A 120-g cube of ice at 0°C is dropped into 1.0 kg of water that was originally at 75°C. What is the final temperature of the water after the ice has melted?

Answer 1

here,

the mass of ice , m1 = 120 g = 0.12 kg

mass of water , m2 = 1 kg

specific heat of water , Cw = 4186 J/kg degree C

latent heta of fusion , Lf = 3.34 * 10^5 J/kg

let the final temprature of water be Tf

using conservation of heat energy

m1 * ( Lf + Cw * ( Tf - 0)) = m2 * Cw * ( 70 - Tf)

0.12 * ( 3.34 * 10^5 + 4186 * ( Tf - 0)) = 1 * 4186 * ( 70 - Tf)

solving for Tf

Tf = 53.95 degree C

the final temprature of water is 53.95 degree C