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Question

(B) The total heat to be removed $(Q)$ consists of two parts: cooling the water from $30^{\circ}C$ to $0^{\circ}C$ and freezing the water at $0^{\circ

Vedclass · Accepted Answer

$1925$

Vedclass · Answer

(B) The total heat to be removed $(Q)$ consists of two parts: cooling the water from $30^{\circ}C$ to $0^{\circ}C$ and freezing the water at $0^{\circ}C$ into ice.$1$. Heat to cool water: $Q_1 = m \cdot c \cdot \Delta T = 15 \ kg 	imes 4200 \ J/(kg \cdot K) 	imes 30 \ K = 1,890,000 \ J$.$2$. Heat to freeze water: $Q_2 = m \cdot L_f = 15 \ kg 	imes 3.36 	imes 10^5 \ J/kg = 5,040,000 \ J$.Total heat $Q = Q_1 + Q_2 = 1,890,000 + 5,040,000 = 6,930,000 \ J$.Time $t = 1 \ hour = 3600 \ s$.Power $P = Q / t = 6,930,000 / 3600 = 1925 \ W$.

The power of a refrigerator that can make $15 \ kg$ of ice at $0^{\circ}C$ from water at $30^{\circ}C$ in one hour is: (in $W$)

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