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

Step-by-step 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* 1000* 80\\\Rightarrow Q=24000\ cal\ or\ 24\ kcal`

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