Water of volume 2 liter in a container is heated with a coil of 1 kW at 27^\circ C . The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27^\circ C to $77^\circ \mathrm{C}$ [Given specific heat of water is 4.2 kJ/kg ]
Text Solution
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Heat gained by the water = (Heat supplied by the coil) – (Heat dissipated to environment)
⇒ ⇒ $mc \Delta \theta = P_{\mathrm{Coil}} t - P_{\mathrm{Loss}} t$
⇒ ⇒ $2 \times 4.2 \times 10^3 \times (77 - 27) = 1000t - 160 t$
$\Rightarrow t = \frac{4.2 \times 10^{5}}{840} = 500 \text{ sec} = 8 \text{ min } 20 \text{ sec}$
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