How is the Energy for Water Evaporation Calculated in a Student Experiment?

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The discussion focuses on calculating the energy required for water evaporation in a student experiment involving two measuring cylinders at different temperatures. The specific latent heat of vaporization for water is given as 2300 J/g, and the density is 1.0 g/cm³. The student correctly identifies that the volume difference in cylinder B, which is 20 cm³, corresponds to 20 g of water. However, the calculation must include the energy needed to heat the water to its boiling point before evaporation occurs. Thus, the total energy calculation involves both the heating and the latent heat of vaporization.
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I need help for number 2(b)

A student investigates the evaporation of water. He pours 100 cm3 of water into measuring
cylinder A and 100 cm3 of water into measuring cylinder B. Measuring cylinder A is kept
at 40 °C and B is kept at 80 °C in the same part of the laboratory. Fig. 2.1 shows the two
measuring cylinders after 3 days.

The specific latent heat of vaporisation of water is 2300 J / g and the density of water is
1.0 g / cm3. During the three days, the water level in B drops from the 100 cm3 mark to
the level shown in Fig. 2.1. Calculate the energy used to evaporate water from B during
the three days.








Volume = (80-60) = 20 cm3

1 cm3 = 1 g
20 cm3 = 20 g

Q= ml
Q= 20/1000 x 2300
= 46(Not the right answer)


Thanks.
 
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I see 2 difficulties.
First, the 20 grams should not be converted to kg because the latent heat "2300 J / g" is in grams, not kg.

Second, I think you have to add the heat required to warm the 20 g up to 100 degrees.
 
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