Question

How much energy in watt-hour is required for an electric heater to convert 2 kg of ice at $0^{\circ }C$ to water ${25}^{\circ }C$. Assume the water equivalent of the given heater to be 10 g. (Assume that initially the heater and the ice are in thermal equilibrium).

Answer 1

Given: An electric heater to convert 2 kg of ice at $0^{\circ }C$ to water ${25}^{\circ }C$. Assume the water equivalent of the given heater to be 10 g. (Assume that initially the heater and the ice are in thermal equilibrium).

To find the amount of energy required by the electric heater for the conversion.

Solution:

The quantity of heat Q supplied or given out during a change of state is given by:

$Q=mL$

where $m$ is the mass in kilograms and $L$ is the specific latent heat.

So here the mass of ice =2kg

and specific latent heat of fusion of ice, $L=334kJ/kg$

So,

$Q_1=2\times 334=668kJ....\left(i\right)$

This is the Latent heat needed to change ice at $0°C$ into water at $0°C$

Heat energy needed to change the temperature of water from 0°C (i.e. melting point) to $25°C$, is

$Q=mc(t_2-t_1)$,

where $m$ is mass of ice, $c=4.2kJ/\left(k{g}^{\circ }C\right)$ is specific heat capacity of water and $t_2-t_1={25}^{\circ }C$ is change in temperature

According to the given criteria,

Hence,

$Q_2=2\times 4.2\left(25\right)=210kJ$

Total heat = Heat required to convert 2 kg of ice to 2 kg of water at $0°C$ + Heat required to convert 2 kg of water at $0°C$ to 2 kg of water at $25°C$

i.e., $668+210=878kJ=243.9Wh$

Hence the energy required for an electric heater for particular conversion is 243.9Wh.