JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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).
- A \(301\)
- B \(302\)
- C \(300\)
- D \(303\)
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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