22. How much heat energy is liberated when 20 kg of water
at 0 °C freezes to form ice at 0 °C? Take latent heat of
ice as 336 kJ kg-1
(Ans. 6720 kJ)
Answers
Explanation:
Total heat = Heat required to convert 2kg of ice
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kg
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJ
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-k
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-kΔt(temperature difference) =20℃
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-kΔt(temperature difference) =20℃Therefore,
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-kΔt(temperature difference) =20℃Therefore,Heat required=2×334+2×4.187×(20−0)
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-kΔt(temperature difference) =20℃Therefore,Heat required=2×334+2×4.187×(20−0) =835.48 kJ
Total heat = Heat required to convert 2kg of iceto 2kg of water at 0℃ +Heat required to convert 2kg of water at 0℃ to 2kg of water at 20℃Heat=mhfg+mCp\Delta t$$Here, m(mass of ice)=2kghfg(latent heat of fusion of ice)=334kJCp of water(specific heat)=4.187 kJ/kg-kΔt(temperature difference) =20℃Therefore,Heat required=2×334+2×4.187×(20−0) =835.48 kJTherefore, to melt 2kg of ice 835.48kJ of heat is required.