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Calculate the quantity of heat required to convert 10^(-2)kg of ice at 0^(@)C to steam at 100^(@)C. (Given L_("ice")=0.335xx10^(6)Jkg^(-1),L_("steam")=2.26xx10^(6)Jkg^(-1), Specific heat of water = 4186Jkg^(-1)K^(-1)).

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Quantity of <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> required to <a href="https://interviewquestions.tuteehub.com/tag/convert-425734" style="font-weight:bold;" target="_blank" title="Click to know more about CONVERT">CONVERT</a> ice into water at `0^(@)C(Q_(1))=mL_("ice")=10^(-2)xx0.335xx10^(6)=3350J` <br/> Quantity of heat required to raise the temperature of water from `0^(@)C" to "100^(@)C`. <br/> `(Q_(2))=mS_(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)(t_(2)-t_(1))=10^(-2)xx4186xx100=4186J` <br/> Quantity of heat required to convert water into steam at `100^(@)C` <br/> `(Q_(3))=mL_("steam")=10^(-2)xx2.26xx10^(6)=22600J` <br/> Total amont of heat required `(<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>)=Q_(1)+Q_(2)+Q_(3)` <br/> = 3350 + 4186 + 22600 = 30136 J.</body></html>


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