A) 716 cal
B) 500 cal
C) 180 cal
D) 100 cal
Correct Answer: C
Solution :
Heat required to convert\[0{}^\circ C\]of ice in water \[=mL\] (L = latent heat) Now, heat required to rise the temperature of water from\[0{}^\circ C\]to\[100{}^\circ C\]. \[=ms\Delta t\] where\[s=\]specific heat.\[\Delta t=\]temperature difference. Total heat\[=mL+ms\Delta t\] \[=1\times 80+1\times 1\times (100-0)\] \[=180cal\]You need to login to perform this action.
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