A) 3045 J
B) 6056 J
C) 721 J
D) 616 J
Correct Answer: A
Solution :
Heat energy required to change \[-{{10}^{o}}C\]ice into \[{{0}^{o}}C\]ice \[{{Q}_{1}}=m{{C}_{1}}\Delta t\] Heat energy required to change \[{{0}^{o}}C\] ice into water \[{{Q}_{2}}=m{{L}_{f}}\] Heat energy required to change \[{{0}^{o}}C\] water mar \[{{100}^{o}}C\]water \[{{Q}_{3}}=m{{C}_{w}}\Delta t'\] Heat energy required to change \[{{100}^{o}}C\] water into \[{{100}^{o}}C\] steam \[{{Q}_{4}}=m{{L}_{s}}\] \[\therefore \]Total heat energy required \[Q=m{{C}_{1}}\,\Delta t+m{{L}_{f}}+m{{C}_{w}}\Delta t'+m{{L}_{4}}\] \[=1(0.5\times 10+80+1\times 100+540)cal\] \[=(5+80+100+540)\times 4.2J\] \[=3045J\]
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