Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C?

2017-07-19 9:06 am

the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 17.7 g of steam at 232°C is condensed, cooled, and frozen to ice at -50.°C.

回答 (1)

胡雪8°

2017-07-19 10:14 am

✔ 最佳答案

Molar mass of water (H₂O) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of water = (17.7 g) / (18.0 g/mol) = 17.7/18.0 mol

For the change of 232°C steam to 100°C steam :
Heat lost = m c ΔT = 17.7 × 2.03 × (232 - 100) J = 4743 J = 4.743 kJ

For the change of 100°C steam to 100°C water :
Heat lost = n Lv = (17.7/18.0) × 40.6 kJ = 39.923 kJ

For the change of 100°C water to 0°C water :
Heat lost = m c ΔT = 17.7 × 4.18 × (100 - 0) = 7399 J = 7.399 kJ

For the change of 0°C water to 0°C ice :
Heat lost = n Lf = (17.7/18.0) × 6.02 kJ = 5.920 kJ

For the change of 0°C ice to -50°C ice :
Heat lost = m c ΔT = 17.7 × 2.06 × [0 - (-50)] J = 1823 J = 1.823 J

Total heat lost = 4.743 + 39.923 + 7.399 + 5.920 + 1.823 kJ = 59.8 kJ (to 3 sig. fig.)

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