Thermal Properties of Matter

JEE Main level, single correct, medium

12 of 126

Question 12 of 126

Time taken by a 836 W836 \mathrm{~W} heater to heat one litre of water from 10∘C10^{\circ} \mathrm{C} to 40∘C40^{\circ} \mathrm{C} is

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Answer: (C) 150 s150 \mathrm{~s}

Let t\mathrm{t} be the time taken, then
836×t4.2=1000×1×(40−10)[ using Q=mst]⇒t=150 sec. \begin{aligned} & \frac{836 \times t}{4.2}=1000 \times 1 \times(40-10)[\text { using } Q=m s t] \\ & \Rightarrow t=150 \text { sec. } \end{aligned}

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