A) 3100
B) 3200
C) 3600
D) 4200
Correct Answer: C
Solution :
Total required heat (Q)= heat required to convert ice into water at\[0{}^\circ C\]\[({{Q}_{1}})+heat\]required to increase the temperature of water up to\[100{}^\circ C({{Q}_{2}})+heat\]required to convert water into steam at\[100{}^\circ C({{Q}_{3}})\]. \[\therefore \] \[{{Q}_{1}}=mL\] \[=5\times 80=400\,cal\] \[{{Q}_{2}}=ms\Delta t\] \[=5\times 1\times (100-0)\] \[=500\,cal\] \[{{Q}_{3}}=mL=5\times 540\] \[=2700\,cal\] Thus, \[Q=400+500+2700\] \[=3600\text{ }cal\]
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