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Water of mass 100 g at a temperature of 100 °C is converted into steam at 100 °C. The specific
latent heat of vaporisation of water is 2300 J/g.
How much thermal energy is absorbed by the water?
0 230 000
0 23 000 000 J
0 230 J
O 23 J

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Answer:

The thermal energy absorbed by the water is 230,000 J.

Explanation:

Given;

mass of the water, m = 100 g

The specific latent heat of vaporization of water, L = 2300 J/g

temperature of the steam, θ = 100 °C

The thermal heat capacity or thermal energy absorbed by the water is calculated as follows;

H = mL

H = 100g x 2300J/g

H = 230,000 J

Therefore, the thermal energy absorbed by the water is 230,000 J.

Therefore

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