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A refrigerator with an average cop of 2.8 used to cool a well-insulated container whose contents are equivalent to 10 kg of water, from 40'C to 10'C. The refrigerator consumes 400 W of electric power. Considering the specific heat of water as 4.2 kj/kgC, the time required to cool the water is ? my answer is 19 min.is it true ?

Respuesta :

Answer:

Yes, it is true

Explanation:

The computation of the time required to cool the water is shown below:

GIven that

Average cop = 2.8

Therefore the cooling effect is

= 2.8 × 400 W

= 1,120 W

ANd, the specific heat of water is 4.2 kj/kg°C i.e 4,200 j/kg°C

Now we assume the time is t

As we know that

[tex]1,120 \times t = 10\times 4,200 \times (40 - 10)\\\\ therefore\\\\ t = \frac{10\times4,200\times30}{1,120}[/tex]

t = 1,125 seconds

So, it would be 19 minutes

Therefore it is true

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