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

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

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

c)
`m=66.77\ kg`

Step-by-step 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=(\bar E)/(365) * 3600`

`E=(446000* 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=(Desired\ effect\ heat)/(Energy\ supplied )`

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

`6.3=(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=(E_r)/(c_w* \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*10^5\ J.kg^(-1)`

`m=(27713095.89)/(4186* 19.6+3.33* 10^5)`

`m=66.77\ kg`