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 into water = mi (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 \[=mL+ms\,\Delta t\] \[\begin{align} & =1\times 80+1\times 1\times (100-0) \\ & =180\,cal \\ \end{align}\]You need to login to perform this action.
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