Question

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C. If the watermelons are initially at 28°C, determine how long it will take for the refrigerator to cool them? The watermelons can be treated as water whose specific heat is 4.2 kJ/kg·°C. Is your answer realistic or optimistic?

Answer 1

6222.22 sec

Step-by-step explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature
`T_1=8^(\circ)C`

`T_2=28^(\circ)C`

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

`Q=mc_pdt=5* 10* 4.2* (28-8)=4200 KJ`

We know that
`COP=(Q_1)/(W)`

`1.5=(Q_1)/(450)`

`Q_1=675` W=0.675 KW

so time required to cool the watermelon is

`t=(Q_1)/(Q_2)=(4200)/(0.675)=6222.22 sec`