Question

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 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.

Answer 1

Answer:

$\Delta t = 5866.667\,s\,(97.778\,m)$

Explanation:

The specific heat for watermelon above freezing point is $3.96\,\frac{kJ}{kg\cdot K}$. The heat liberated by the watermelon to cool down to 8°C is:

$Q_{cooling} = (5)\cdot (10\,kg)\cdot (3.96\,\frac{kJ}{kg\cdot K} )\cdot (20\,K)$

$Q_{cooling} = 3960\,kJ$

The heat absorbed by the household refrigerator is:

$\dot Q_{L} = COP\cdot \dot W_{e}$

$\dot Q_{L} = 1.5\cdot (0.45\,kW)$

$\dot Q_{L} = 0.675\,kW$

Time needed to cool the watermelons is:

$\Delta t = \frac{Q_{cooling}}{\dot Q_{L}}$

$\Delta t = \frac{3960\,kJ}{0.675\,kW}$

$\Delta t = 5866.667\,s\,(97.778\,m)$