A mixture consists of 70 g of ice and 60 g of liquid water, both at 0$\degree$C. (The latent heat of fusion of water is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C$\degree$.) The amount of heat that must be added to melt all of the ice is about
A) 4000 cal.
B) 5600 cal.
C) 21,600 cal.
D) 32,400 cal.

Question

Quizplus · Accepted Answer

To melt the ice, we need to add enough heat to bring it to its melting point (0°C), then to melt it completely, and finally, add enough heat to raise the temperature of the resulting liquid water to 0°C.
The heat required to raise the temperature of the ice to 0°C is:
Q1 = m1 × Cice × ΔT1
Where m1 is the mass of ice, Cice is the specific heat capacity of ice, and ΔT1 is the change in temperature from -18°C to 0°C.
Q1 = 70 g × 0.5 cal/g°C × 18°C = 630 cal

The heat required to melt the ice is:
Q2 = m1 × Lf
Where Lf is the latent heat of fusion of water.
Q2 = 70 g × 80 cal/g = 5600 cal

The heat required to raise the temperature of the resulting liquid water to 0°C is:
Q3 = m2 × Cwater × ΔT2
Where m2 is the mass of liquid water, Cwater is the specific heat capacity of water, and ΔT2 is the change in temperature from 0°C to 0°C.
Q3 = 60 g × 1.0 cal/g°C × 0 = 0 cal
(Note that ΔT2 is 0 because the temperature doesn't change.)

Therefore, the total heat required is:
Qtot = Q1 + Q2 + Q3 = 630 + 5600 + 0 = 6230 cal
The closest answer is B) 5600 cal.

A Mixture Consists of 70 G of Ice and 60 °\degree°

A Mixture Consists of 70 G of Ice and 60 $\degree$