Solving for Latent Heat of Vaporization: A Student's Journey

[DarkPhoenix]

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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.

 

[Delphi51]

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.

 

[tomcue]

Hello there,

To solve for the energy used to evaporate water from cylinder B during the three days, we need to first calculate the mass of water that has evaporated. We know that the volume of water in cylinder B has decreased from 100 cm3 to 20 cm3, so the mass of water that has evaporated is:

m = 100 cm3 - 20 cm3 = 80 cm3
1 cm3 of water has a mass of 1 g, so the mass of water evaporated is 80 g.

Now, we can use the formula Q = ml to calculate the energy used to evaporate this amount of water. We know that the specific latent heat of vaporization of water is 2300 J/g, so the energy used is:

Q = 80 g x 2300 J/g = 184,000 J

Therefore, the energy used to evaporate water from cylinder B during the three days is 184,000 J. I hope this helps. Let me know if you have any further questions. Good luck with your experiment!