Question

A Carnot engine work as refrigerator in between 0^(@)C and 27^(@)C. How much energy is needed to freeze 10 kg ice at 0^(@)C/

Answer 1

Answer:

Work required to freeze 10 kg water into ice is given as

$W = 3.31 \times 10^5 J$

Explanation:

As we know that water at 0 degree is converted into ice at 0 degree

So here heat is removed from it and it is given as

$Q = mL$

now we know that

m = 10 kg

$L = 3.35 \times 10^5 J/kg$

now we have

$Q = 10 (3.35 \times 10^5)$

$Q_2 = 3.35 \times 10^6 J$

now we know that COP of refrigerator is given as

$\frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$

$\frac{3.35 \times 10^6}{W} = \frac{273}{300 - 273}$

$W = 3.31 \times 10^5 J$

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Topic : COP of refrigerator

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