0.3 kg * .75 = 0.225 kg of ice
first bring ALL the ice up 8 degrees C to 0.0 C
heat in = 0.300 * specific heat of ice in Joules/kg deg C * 8
then
melt 0.225 kg ONLY of the ice at 0 .0 C
heat in = 0.225 * heat of fusion of water
add those heats in Joules
heat of fusion of water Lf= 3.33* 10^5 J/Kg
specific heat of ice = Cw = 2093 J/Kg degC
A 0.300-kg of ice is initially at temperature of −8.0 °C. How much heat is required to melt three-quarters the mass of the ice only?
1 answer
1 answer