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# Time taken by a $836\: W$ heater to heat one liter of water from $10^{\circ}C\: to\: 40^{\circ}C$ is

$\begin {array} {1 1} (A)\;50s & \quad (B)\;100s \\ (C)\;150s & \quad (D)\;200s \end {array}$

Let $t$ be the time taken
Then heat energy in calories is given by
$\large\frac{836 \times t}{4.2}$ $= 1000 \times 1 \times (40 – 10)$ [using $Q = mst$ (from heat transfer)]
$\Rightarrow t = 150s$
Ans : (C)

edited Sep 10, 2014