A) 3045 J
B) 6056 J
C) 721 J
D) 616 J
Correct Answer: A
Solution :
Ice (?10°C) converts into steam as follows (ci = Specific heat of ice, cW = Specific heat of water) Total heat required \[Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] Þ \[Q=1\times 0.5(10)+1\times 80+1\times 1\times (100-0)+1\times 540\] \[=725\,cal\] Hence work done \[W=JQ=4.2\times 725=3045\,J\]You need to login to perform this action.
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