Question

Calculate the heat change in calories for melting of 0.30 kg of water at 0*C. The
heat of fusion for water is 80 cal/g. The heat of vaporization of water is 540 cal/g.
The specific heat capacity of water is 1.00 cal/g*C.

Answer 1

Answer: 24 kcal

Explanation:

Given

Mass of water $m=0.3\ kg$

Temperature of water $T_1=0^{\circ}$

Heat of fusion $L_f=80\ cal/g$

Heat of vaporization $L_v=540\ cal/g$

Specific heat of water $c=1\ cal/g.^{\circ}C$

Heat require to melt the ice is

$\Rightarrow Q=mL_f\\\Rightarrow Q=0.3\times 1000\times 80\\\Rightarrow Q=24000\ cal\ or\ 24\ kcal$

Thus, 24 kcal of heat is required to melt 0.3 kg of ice.