Water of volume 2, L in a closed container is | vedclass

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(A) Given:Volume of water $V = 2\, L$,so mass $m = 2\, kg$ (since density of water is $1\, kg/L$).Power of coil $P_{in} = 1000\, J/s$.Rate of heat los

Vedclass · Accepted Answer

$8\, min\, 20\, s$

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(A) Given:Volume of water $V = 2\, L$,so mass $m = 2\, kg$ (since density of water is $1\, kg/L$).Power of coil $P_{in} = 1000\, J/s$.Rate of heat loss $P_{out} = 160\, J/s$.Temperature change $\Delta T = 77\, ^\circ C - 27\, ^\circ C = 50\, ^\circ C$.Specific heat capacity of water $c = 4.2\, kJ/kg\cdot K = 4200\, J/kg\cdot K$.Net power supplied to the water $P_{net} = P_{in} - P_{out} = 1000\, J/s - 160\, J/s = 840\, J/s$.Total heat required $Q = m \cdot c \cdot \Delta T = 2\, kg 	imes 4200\, J/kg\cdot K 	imes 50\, K = 420,000\, J$.Time required $t = \frac{Q}{P_{net}} = \frac{420,000\, J}{840\, J/s} = 500\, s$.Converting to minutes: $500\, s = 8\, min\, 20\, s$.

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