Question

100 g of solid ice at 0 degree Celcius is added to 500 g of liquid water at 90 degree celcius. What is the final temperature? What initial water temperature is required so that 10 g of ice remain once equilibrium has been reached?

Answer 1

here,

mass of ice , m1 = 100 g = 0.1 kg

mass of water , m2 = 500 g = 0.5 kg

Ti = 90 degree

let the final temprature be Tf

heat gained by ice = heat lost by water

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

0.1 * ( 334000 + 4186 * ( Tf - 0)) = 0.5 * 4186 * ( 90 - Tf)

solving for Tf

Tf = 61.7 degree

the final temprature is 61.7 degree

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let the initial temprature be Ti

for the 10 g ice not melt

final temprature is 0 degree C

heat gained by ice = heat lost by water

(m1 - 0.01) * ( Lf) = m2 * Cw * ( Ti - 0)

(0.1 - 0.01) * 334000 = 0.5 * 4186 * Ti

Ti = 14.4 degree

the initial temprature of water is 14.4 degree C