A) 1028 ealorie
B) 624 calorie
C) 745 calorie
D) 810 calorie
Correct Answer: C
Solution :
Heat required to raise the temperature of ice from\[-50{}^\circ C\]to\[0{}^\circ C\]is \[{{Q}_{1}}=ms\Delta t=1\times 0.5\times 50\]=25 calorie Heat required to liquify 1 g ice at\[0{}^\circ C\]to g water at\[0{}^\circ C\]is: \[{{Q}_{2}}=mL\] \[=1\times 80=80\text{ }calorie\] Heat required to boil the water from\[0{}^\circ C\]to \[100{}^\circ C\] \[{{Q}_{3}}=ms\Delta t\] \[=1\times 1\times 100\] \[=100\text{ }calorie\] Heat required to change 1 g of water to vapour is \[{{Q}_{4}}=mL\] \[=1\times 540\] \[=540\text{ }calorie\] \[\therefore \] Total heat required in whole process \[Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[=25+80+100+540\] \[=745\text{ }calorie\]
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