45g of water at 50oC in a beaker is cooled when 50g of copper at 18oC is added to it. the contents are stirred till a final constant temperature is reached .the specific heat capacity of copper is 0.39Jkg-1K-1.state the assumption used.

Heat gained by copper=m×C×θ=50×0.39×(T-18)Heat lost by water=45×4.2×(50-T)Heat lost=Heat gained45×4.2×(50-T)=50×0.39×(T-18)9450-189 T=19.5 T-351208.5 T=9801Or T=470C (Approx)So the fnal temperature is 470C (Approx).

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