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

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

Step-by-step explanation:

The specific heat for watermelon above freezing point is
`3.96\,(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\,(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 = (Q_(cooling))/(\dot Q_(L))`

`\Delta t = (3960\,kJ)/(0.675\,kW)`

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