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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
  • JEE Mains 2023
  • Medium
  • Numerical

A water heater of power \(2000\,W\) is used to heat water. The specific heat capacity of water is \(4200\,J\,kg ^{-1}\, K ^{-1}\). The efficiency of heater is \(70 \%\). Time required to heat \(2\,kg\) of water from \(10^{\circ}\,C\) to \(60^{\circ}\,C\) is \(..........s\). (Assume that the specific heat capacity of water remains constant over the temperature range of the water).

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Answer & Solution

Correct Answer

(C) \(300\)

Step-by-step Solution

Detailed explanation

\(Q = mc\Delta T = 2 \times 4200 \times (60 - 10) = 420000\,J\)

\(t = \frac{Q}{\eta P} = \frac{420000}{0.70 \times 2000} = 300\,s\)

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