Chemistry heat change?

2017-10-18 12:05 pm
更新1:

Calculate the heat change in calories for condensation of 13.0 g of steam at 100∘C. Calculate the heat change in joules for condensation of 7.10 g of steam at 100∘C. Calculate the heat change in kilocalories for vaporization of 42 g of water at 100∘C. Calculate the heat change in kilojoules for vaporization of 9.00 kg of water at 100∘C.\ Can't figure this out. Please and thanks

回答 (1)

2017-10-18 12:46 pm
Calculate the heat change in calories for condensation of 13.0 g of steam at 100°C.

Latent heat of vaporization of water at 100°C = 540 cal/g

Heat is released in condensation.
Heat change = -(13.0 g) × (540 cal/g) = -7020 J


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Calculate the heat change in joules for condensation of 7.10 g of steam at 100°C.

Latent heat of vaporization of water at 100°C = 2260 J/g

Heat is released in condensation.
Heat change = -(7.10 g) × (2260 J/g) = -16000 J


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Calculate the heat change in kilocalories for vaporization of 42 g of water at 100°C.

Latent heat of vaporization of water at 100°C = 540 cal/g

Heat is absorbed in vaporization.
Heat change = (42 g) × (540 cal/g) = 22700 cal = 22.7 kcal


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Calculate the heat change in kilojoules for vaporization of 9.00 kg of water at 100°C.

Latent heat of vaporization of water at 100°C = 2260 kJ/kg

Heat is absorbed in vaporization.
Heat change = (9.00 kg) × (2260 kJ/kg) = 20300 kJ


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