A student attempted to measure the specific latent heat of vaporisation of water.

She took the following readings:

First balance reading (g) = 581
Second balance reading (g) = 526
First joulemeter reading (kJ) = 195
Second joulemeter reading (kJ) = 327

Use these results to determine the specific latent heat of vaporisation of water.

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oeerivona oeerivona · Answer 1 · 2021-01-08T07:52:56+01:00

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g