Question

A 0.20 kg ice cube at 0.0° C has sufficient heat added to it to cause total melting, and the resulting water is heated to 33 °C. How much heat is added? (For water L_f= 334 kJ/kg and L_v= 2257 kJ/kg.)

Answer 1

We'll need to calculate the energy required to melt the water and heat it to 33°C

To melt the ice, you'll need to use the equation Q = Lf X m
Q= Energy in kilojoules
Lf= latent heat of fusion of water= 334
m= mass of water

Q= 334 X 0.2=66.8 kJ


To heat it to 33°C, you need to use Q=mc (delta T)
Q= Energy in kilojoules
m=mass of water = 0.2
c=specific heat capacity of water in kJkg^-1°C^-1 = 4.2
delta T= change in temperature= 33
Q= 0.2 X 4.2 X 33 = 27.72 kJ

Total energy= 66.8 +27.72= 94.52 kJ