A water cooler of storage capacity 120 litres | vedclass

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(B) Total heat generated by the device in $t = 3 	ext{ hours} = 3 	imes 3600 	ext{ s} = 10800 	ext{ s}$ is:$Q_{	ext{gen}} = P_{	ext{device}} 	i

Vedclass · Accepted Answer

$2067$

Vedclass · Answer

(B) Total heat generated by the device in $t = 3 	ext{ hours} = 3 	imes 3600 	ext{ s} = 10800 	ext{ s}$ is:$Q_{	ext{gen}} = P_{	ext{device}} 	imes t = 3000 	ext{ W} 	imes 10800 	ext{ s} = 3.24 	imes 10^7 	ext{ J}$.The heat absorbed by the $120 	ext{ litres}$ $(120 	ext{ kg})$ of water as its temperature rises from $10^{\circ}	ext{C}$ to $30^{\circ}	ext{C}$ is:$Q_{	ext{water}} = m \cdot c \cdot \Delta T = 120 	ext{ kg} 	imes 4200 	ext{ J kg}^{-1} 	ext{K}^{-1} 	imes (30 - 10) 	ext{ K} = 120 	imes 4200 	imes 20 = 1.008 	imes 10^7 	ext{ J}$.The heat that must be removed by the cooler is:$Q_{	ext{cooler}} = Q_{	ext{gen}} - Q_{	ext{water}} = 3.24 	imes 10^7 	ext{ J} - 1.008 	imes 10^7 	ext{ J} = 2.232 	imes 10^7 	ext{ J}$.The minimum power $P$ of the cooler is:$P = \frac{Q_{	ext{cooler}}}{t} = \frac{2.232 	imes 10^7 	ext{ J}}{10800 	ext{ s}} \approx 2066.67 	ext{ W} \approx 2067 	ext{ W}$.

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