Question

A copper block of mass 4 kg is heated in a fumance to a temperature $ \text{425}{{\,}^{\text{o}}}\text{C} $ and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper $ =500\,J\,k{{g}^{-1}}-{{\,}^{o}}{{C}^{-1}} $ and heat of fusion of ice $ =336\,kJ\,k{{g}^{-1}} $ ):

• A. 0.5 kg
• B. 1 kg
• C. 1.5 kg
• D. 2.5 kg

Answer 1

Answer: (D)

Fall in temperature of copper block when it is placed on the ice block $ =\Delta t=425-0=425{{\,}^{o}}C $ Heat lost by copper block when it is placed on the ice block. $ {{Q}_{1}}={{m}_{1}}s\Delta T $ $ =4\times 500\times 425=850\,kJ $ Heat gained by ice in melting into $ {{m}_{2}}\,kg $ of water. $ {{Q}_{2}}={{m}_{2}}L $ $ ={{m}_{2}}\times 336 $ $ =336\,{{m}_{2}}\,kJ $ According to calorimetry principle, heat lost = heat gained i.e., $ 850=336\,{{m}_{2}} $ $ \therefore $ $ {{m}_{2}}=\frac{850}{336}=2.5\,kg $