Please help with this chemistry question..?
2017-11-08 1:30 pm
Assuming no work is performed, estimate how much heat is required to take 18.4 kg of liquid water from 40 degrees C to steam at 160 degrees C. The specific heat of liquid water is 4.184 J/(g K), the specific heat of steam is 2.08 J/(g K), the latent heat of fusion of water is 6.02 kJ/mol, and the latent heat of vaporization of water is 40.7 kJ/mol
回答 (1)
2017-11-08 2:48 pm
Molar mass of water (H₂O)
= (1.0×2 + 16.0) g/mol
= 18.0 g/mol
= 0.0180 kg/mol
Heat required for 40°C liquid water to 100°C liquid water
= (18.4 kg) × [4.184 kJ/(kg °C)] × [(100 - 40) °C]
= 18.4 × 4.184 × 60 kJ
Heat required for 100°C liquid water to 100°C steam
= [(1.84 kg) / (0.0180 kg/mol)] × (407 kJ/mol)
= 1.84 × 407 / 0.0180 kJ
Heat required for 100°C steam to 160°C steam
= (18.4 kg) × [2.08 kJ/(kg °C)] × [(160 - 100) °C]
= 18.4 × 2.08 × 60 kJ
Total heat required
= (18.4 × 4.184 × 60) + (1.84 × 407 / 0.0180) + (18.4 × 2.08 × 60) kJ
= 423000 kJ
= (1.0×2 + 16.0) g/mol
= 18.0 g/mol
= 0.0180 kg/mol
Heat required for 40°C liquid water to 100°C liquid water
= (18.4 kg) × [4.184 kJ/(kg °C)] × [(100 - 40) °C]
= 18.4 × 4.184 × 60 kJ
Heat required for 100°C liquid water to 100°C steam
= [(1.84 kg) / (0.0180 kg/mol)] × (407 kJ/mol)
= 1.84 × 407 / 0.0180 kJ
Heat required for 100°C steam to 160°C steam
= (18.4 kg) × [2.08 kJ/(kg °C)] × [(160 - 100) °C]
= 18.4 × 2.08 × 60 kJ
Total heat required
= (18.4 × 4.184 × 60) + (1.84 × 407 / 0.0180) + (18.4 × 2.08 × 60) kJ
= 423000 kJ
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171108053029AAJMxio