Question

A freezer has a coefficient of performance of 6.30. It is advertised as using 446 kWh/yr. Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour.
(a) On average, how much energy does it use in a single day?
(b) On average, how much energy does it remove from the refrigerator in a single day?
(c) What maximum mass of water at 19.6°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 105 J/kg, and its specific heat is 4186 J/kg · °C.)

Answer 1

Answer:

a) $E\approx4398904.1\ J$ per day

b) $E_r=27713095.89\ J\ per\ day$

c) $m=66.77\ kg$

Explanation:

Given;

coefficient of performance of the refrigerator, $COP=6.3$

energy used per year of operation of the refrigerator, $\bar E=446\ kWh.yr^{-1}$

a)

Energy used in single day:

$E=\frac{\bar E}{365} \times 3600$

$E=\frac{446000\times 3600}{365}$

$E\approx4398904.1\ J$ per day

b)

Energy removed from the refrigerator in a single day:

We have the energy consumed in 1 day as, $E\approx4398904.1\ J$

so,

$\rm COP=\frac{Desired\ effect\ heat}{Energy\ supplied }$

The desired effect of a refrigerator is to eliminate heat from the evaporator.

$6.3=\frac{E_r}{4398904.1}$

$E_r=27713095.89\ J\ per\ day$

c)

Now the maximum mass of water that can freeze form 19.6°C:

$m=\frac{E_r}{c_w\times \Delta T+L_f}$

where:

$c_w=specific\ heat\ of\ water=4186\ J.kg^{-1}.K^{-1}$

$L_f=latent\ heat\ of\ fusion=33.3\times10^5\ J.kg^{-1}$

$m=\frac{27713095.89}{4186\times 19.6+3.33\times 10^5}$

$m=66.77\ kg$