Time taken by a $836\; W$ heater to heat $1\; litre$ of water from $10^{\circ} C$ to $40^{\circ} C$ is
AIEEE 2004, Medium
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$W = JH$ $ \Rightarrow $ $P \times t = J \times m\,s\,\Delta \theta $
$ \Rightarrow $ $t = \frac{{J \times m \times s\Delta \theta }}{P}$ (For water $1\, litre$ = $1\,kg$)
$ \Rightarrow $ $t = \frac{{4.2 \times 1 \times 1000 \times (40 - 10)}}{{836}} = 150\,sec$
Short Trick : use formula $t = \frac{{4200 \times m \times \Delta \theta }}{P}$
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