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Water of volume $2\, L$ in a closed container is heated with a coil of $1\,kW$. While water is heated, the container loses energy at a rate of $160\, J/s$ . In how much time will the temperature of water rise from $27\,^oC$ to $77\,^oC$ ? (Specific heat of water is $4.2\, kJ/kg$ and that of the container is negligible)
$8\,min\, 20\, s$
$6\,min\, 2\, s$
$7\, min$
$14\, min$
Solution
Fromquestion,
In $1\,sec$ heat gained by water
$ = 1\,K\,W – 160J/s$
$ = 1000\,J/s – 160\,J/s$
$ = 840\,J/s$
Total heat required to raise the temperature of water $\left( {volume\,2L} \right)\,from\,{27^ \circ }c\,to\,{77^ \circ }c$
$ = {m_{water}} \times sp.ht{\rm{\backslash }}\Delta \theta $
$ = 2 \times {10^3} \times 4.2 \times 50$
$\left[ {mass = density \times volume} \right]$
$And,840 \times t = 2 \times {10^3} \times 4.2 \times 50$
$or,\,\,\,\,\,\,\,\,\,t = \frac{{2 \times {{10}^3} \times 4.2 \times 50}}{{840}}$
$500s = 8\,\min \,20s$
