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# An immersion heater is rated $836 walt$ In what time it should heat litre of water from $20^{\circ}$ to $40^{\circ}? J= 4.18 \;J/cal$

$\begin{array}{1 1} 100\;m \\ 10\;m \\ 100\;s \\ 10\;mins \end{array}$

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Solution :
Heat taken by water = $msd \theta$
$\qquad= (1000 \times 1) \times 1 \times (40-20)$
$\qquad= 1000 \times 20 \times 4.81 \;J$
$826 \;t= 1000 \times 20 \times 4.18\;J$
$t= \large\frac{1000 \times 20 \times 4.18}{8.36}$$=100\;s$