Water of volume 2 litre in a container is heated with a coil of 1 kW at $27^{0} C$ . The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from $27^{0} C to 77^{0} C$ ? [Given specific heat of water is 4.2 kJ/kg]

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7 min

8 min 20s

6 min 2 s

14 min

answer is A.

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Detailed Solution

Heat gained by the water = (Heat supplied by the coil)-(Heat dissipated to environment)

$\Rightarrow mc ∆ θ = P_{Coil} t - P_{Loss} t$

$\Rightarrow 2 \times 4.2 \times 10^{3} \times (77 - 27) = 1000 t - 160 t$

$\Rightarrow t = \frac{4.2 \times 10^{5}}{840} = 500 \sec = 8 \min 20 \sec$

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