Physics, asked by harshdeepkaurchahal2, 5 months ago

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

Answered by san1485
2

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.

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