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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?

Respuesta :

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature [tex]T_1=8^{\circ}C[/tex]

[tex]T_2=28^{\circ}C[/tex]

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

[tex]Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ[/tex]

We know that [tex]COP=\frac{Q_1}{W}[/tex]

[tex]1.5=\frac{Q_1}{450}[/tex]

[tex]Q_1=675[/tex] W=0.675 KW

so time required to cool the watermelon is

[tex]t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec[/tex]  

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