Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is (a) 150 s (b) 100 s (c) 50 s (d) 200 s

Answer: (a). Heat needed Q = mcΔ T with m = 1 kg (1 litre of water). Q = 1 × 4180 × 30 = 125400 J t = Q/P = (125400)/(836) = 150 s So the answer is (a) 150 s.

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Calorimetry, Latent Heat & Mechanical Equivalent of Heat

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

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Answer: (a)

Heat needed $Q = mc\Delta T$ with $m = 1\text{ kg}$ (1 litre of water).
$Q = 1 \times 4180 \times 30 = 125400\text{ J}$
$t = \dfrac{Q}{P} = \dfrac{125400}{836} = 150\text{ s}$
So the answer is (a) $150\text{ s}$ .

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