An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0oC to 15.0oC in 100 s. Calculate the specific heat capacity of liquid:
A
3×103Jkg−1K−1
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B
12×103Jkg−1K−1
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C
1.2×103Jkg−1K−1
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D
None of the above
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Solution
The correct option is A3×103Jkg−1K−1 Heat supplied by water is, Q=600×100