A) 0.5 kg
B) 1 kg
C) 1.5 kg
D) 2.5 kg
Correct Answer: D
Solution :
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., \[850=336\,{{m}_{2}}\] \[\therefore \] \[{{m}_{2}}=\frac{850}{336}=2.5\,kg\]You need to login to perform this action.
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