Calculate the heat required to convert 3 ; kg | vedclass

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Question

(A) The total heat $Q$ required is the sum of heat for four processes:$1$. Heating ice from $-12 \; ^{\circ}C$ to $0 \; ^{\circ}C$: $Q_{1} = m \cdot s

Vedclass · Accepted Answer

$9.1 	imes 10^{6} \; J$

Vedclass · Answer

(A) The total heat $Q$ required is the sum of heat for four processes:$1$. Heating ice from $-12 \; ^{\circ}C$ to $0 \; ^{\circ}C$: $Q_{1} = m \cdot s_{	ext{ice}} \cdot \Delta T_{1} = 3 \; kg 	imes 2100 \; J \; kg^{-1} \; K^{-1} 	imes (0 - (-12)) \; K = 75600 \; J$.$2$. Melting ice at $0 \; ^{\circ}C$ to water at $0 \; ^{\circ}C$: $Q_{2} = m \cdot L_{	ext{fusion}} = 3 \; kg 	imes 3.35 	imes 10^{5} \; J \; kg^{-1} = 1005000 \; J$.$3$. Heating water from $0 \; ^{\circ}C$ to $100 \; ^{\circ}C$: $Q_{3} = m \cdot s_{	ext{water}} \cdot \Delta T_{2} = 3 \; kg 	imes 4186 \; J \; kg^{-1} \; K^{-1} 	imes (100 - 0) \; K = 1255800 \; J$.$4$. Converting water at $100 \; ^{\circ}C$ to steam at $100 \; ^{\circ}C$: $Q_{4} = m \cdot L_{	ext{steam}} = 3 \; kg 	imes 2.256 	imes 10^{6} \; J \; kg^{-1} = 6768000 \; J$.Total heat $Q = Q_{1} + Q_{2} + Q_{3} + Q_{4} = 75600 + 1005000 + 1255800 + 6768000 = 9104400 \; J \approx 9.1 	imes 10^{6} \; J$.

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