In: Physics
100 g of solid ice at 0°C is added to 500 g of liquid water at 90°C. What is the final temperature? What is initial water temperature is required so that 10 g of ice remain once equilibrium has been reached?
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