## Answers

Heat needed to convert ice at -11^{o}C to 0^{o}C,

Q_{1} = mc_{ice}T = [0.355*2040*{0-(-11)}] J

= 7966.2 J

Heat needed to convert ice to water at 0^{o}C,

Q_{2} = mL_{f} = (0.355*334*10^{3}) = 118570 J

Heat required to convert water from 0^{o}C to water at 100^{o}C,

Q_{3} = mc_{water}T' = {0.355*4186*(100-0)} J

= 148603 J

Heat required to convert water at 100^{o}C to steam at 100^{o}C,

Q_{4} = mL_{v} = (0.355*2256*10^{3}) = 80088 J

Heat required to convert steam at 100^{o}C to steam at 122^{o}C,

Q_{5} = mc_{steam}T'' = {0.355*1996*(122-100)} J

= 14139.664 J

Thus, total heat absorbed, Q = Q_{1}+Q_{2}+Q_{3}+Q_{4}+Q_{5}

= (7966.2+118570+148603+80088+14139.664)

= 369366.864 J

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