Question

A household refrigerator that has a power input of 450 W and a COP of 1.55 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?Explain

Answer 1

`\Delta t=6021.505\ seconds\\\Delta t=100.35\ min`

This Time is optimistic as calculated are based on ideal case. Considering the real case, refrigerated space will gain heat from surrounding and increase work load and hence requires more time to cool watermelons to required temperature.

Step-by-step explanation:

Given:

Power input=
`\dot W`=450 W

COP=1.55

N=5 watermelons

Mass of each watermelon=10 Kg

Initial Temperature=
`T_i=28^ o C`

Final Temperature=
`T_f=8^ oC`

Specific heat=
`C_p=4.2\ KJ/Kg.C`

Required:

How long it will take for the refrigerator to cool watermelons?

Solution:

Heat Removed from watermelons is:

`Q_R=NmC_p(T_i-T_f)\\Q_R=5*10*4.2(28-8)\\Q_R=4200 KJ`

Rate of heat removed:

`\dot Q_L=COP*\dot W\\\dot Q_L=1.55*450\\\dot Q_L=697.5 W\\\dot Q_L=0.6975\ KW`

Time is Calculated as:

`\Delta t=(Q_L)/(\dot Q_L) \\\Delta t=(4200)/(0.6975)\\ \Delta t=6021.505 seconds\\\Delta t=100.35 min`

This Time is optimistic as calculated are based on ideal case. Considering the real case, refrigerated space will gain heat from surrounding and increase work load and hence requires more time to cool watermelons to required temperature.