Question

is placed in the freezer compartment of a refrigerator with a coefficient of performance of 6.00. How much heat energy is exhausted into the room as the water is changed to ice at -5.00 ∘C?

Answer 1

Answer:

Th heat energy that is exhausted into the room as the water is changed to ice at -5.00 ∘C is H=499.2 kJ.

Explanation:

Incomplete question.

1 L of water at 20.0 is placed in the freezer compartment of a refrigerator with a coefficient of performance of 6.00. How much heat energy is exhausted into the room as the water is changed to ice at -5.00 ∘C?

We first can calculate the total amount of heat that has to be taken from the mass of water. We can divide this as 3 process:

1) Cooling the water from 20 °C to 0 °C

2) Solidification (enthalpy of solidification)

3) Cooling the ice from 0 °C to -5 °C

1) We can calculate the heat as:

$h=c_w\Delta T=4.19\cdot (20-0)=83.8\,kJ/kg$

c_w: specific heat of water

2)  We can look up for the latent heat of solifification (or enthalpy of fusion) of water: 333.6 kJ/kg

3) We can calculate the heat as:

$h=c_i\Delta T=2.108\cdot (0-(-5))=10.5\,kJ/kg$

c_w: specific heat of ice

If we sum the heat extracted in every process, we have:

$H=1\,kg\cdot (83.8\,kJ/kg+333.6\,kJ/kg+10.5\,kJ/kg)=427.9\,kJ$

The coefficient of performance can be written as:

$COP=Q/W=6$

This can be interpreted as the freezer consumes 1 kW for every 6 kW it extract from the compartment (and exhaust to the room).

Then, the heat that is exhausted to the room is equal to the heat extracted from the water and the energy consumed by the refrigeration cycle to operate (work W).

We can write this as:

$H_r=Q+W=Q+Q/COP=427.9+427.9/6=427.9+71.3=499.2\,kJ$