Question

A 1.0 kg block of ice is initially at a temperature of ?5

Answer 1

(a)
Specific heat of ice = 2.11 J/g/deg C
Specific heat of water = 4.184 J/g/deg C (both from Wikipedia)

1 kg of ice requires 2110 J to raise its temperature by 1 deg C

To warm the ice to melting point requires
5 * 2110 = 10550 J

1 kg of ice absorbs 80 kcal or 80 * 4184 J in changing from ice to water at the same temperature
= 334720 J

Total energy required to warm the ice and melt it to water at 0 deg C
= 334720 + 10550 = 345270 J

Total energy added to the ice is 5.7 * 10^5
= 580000 J

Excess energy available to warm the water from 0 deg C
= 580000 - 345270
= 234730 J

This excess energy will warm the water to
234730 / 4184 = 56.1 deg C

(b)
If the energy is increased by a factor, to achieve the same final temperature, the mass must increase by the same factor.

For example, if we supply twice as much heat, half the heat would raise 1 kg to 56.1 deg and the other half of the heat would do the same for a second 1 kg mass of ice.

The mass of ice must be increased by a factor of 5