state ice water steam specific heat j/g. degree C 2.06 4.18 2.03 molar heat of f

Question

state ice water steam specific heat j/g. degree C 2.06 4.18 2.03 molar heat of fusion for water kj/mol 6.02 molar heat of vaporization for water kj/mol 40.6
use the information given above, calculate the energy involvedin the conversion of 291 g of steam at 373k to ice at 243k. be sureto use the correct sign. state ice water steam specific heat j/g. degree C 2.06 4.18 2.03 molar heat of fusion for water kj/mol 6.02 molar heat of vaporization for water kj/mol 40.6 molar heat of fusion for water kj/mol 6.02 molar heat of vaporization for water kj/mol 40.6
use the information given above, calculate the energy involvedin the conversion of 291 g of steam at 373k to ice at 243k. be sureto use the correct sign. state ice water steam specific heat j/g. degree C 2.06 4.18 2.03

Explanation / Answer

Total heat energy released, qT = q(to convert steam at 373 K to water at 373K) +q(tocovert water at 373 K to 273K)                                                     +q(to convert water at 273K to ice at 273K) +q(to convert ice at 273Kto ice at 243K                                                = moles of steam *Hvap + m s(water) T + moles ofwater *Hfus + m s(ice) T                                                = (291g/18g/mol) 40.6 kJ/mol + 291 *4.18 J/g0K *100K + (291g/18g/mol)*6.02 kJ/mol                                                     + 291g * 2.06 J/g 0K * 30K                                                = 656 kJ + 121.6 kJ + 97.32 kJ + 17.98 kJ                                                = 892.9 kJ Here energy is released so, qT =- 892.9 kJ
  

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