Answer:
[tex]\Delta t = 5866.667\,s\,(97.778\,m)[/tex]
Explanation:
The specific heat for watermelon above freezing point is [tex]3.96\,\frac{kJ}{kg\cdot K}[/tex]. The heat liberated by the watermelon to cool down to 8°C is:
[tex]Q_{cooling} = (5)\cdot (10\,kg)\cdot (3.96\,\frac{kJ}{kg\cdot K} )\cdot (20\,K)[/tex]
[tex]Q_{cooling} = 3960\,kJ[/tex]
The heat absorbed by the household refrigerator is:
[tex]\dot Q_{L} = COP\cdot \dot W_{e}[/tex]
[tex]\dot Q_{L} = 1.5\cdot (0.45\,kW)[/tex]
[tex]\dot Q_{L} = 0.675\,kW[/tex]
Time needed to cool the watermelons is:
[tex]\Delta t = \frac{Q_{cooling}}{\dot Q_{L}}[/tex]
[tex]\Delta t = \frac{3960\,kJ}{0.675\,kW}[/tex]
[tex]\Delta t = 5866.667\,s\,(97.778\,m)[/tex]
