# Question

Hi, any idea why ans is (2)? Thanks a lot.

Setup A has temperature inside 250 C and outside temperature 250 C, so there is no change in temperature hence no condensation on the surface of the container implying there is no change in mass of setup A.

Setup B has temperature inside 50 C and outside temperature 250 C , so the cold inside the container comes in touch with the warmer air, the water vapour in the air condenses into water droplets, forming a layer of water droplets on the surface of the container.  Hence this will increase the mass in set up B.

The layer of oil on top of the water in both setups is to prevent the heat exchanges both ways so that the temperature of water inside the container can retain its original temperature longer for the condensation to take place in setup B.

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Thank you so much sushi88. God bless you 🙏

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My goodness 🙂  So we have to consider the water vapor CONDENSING on the beaker?  Isn’t that an external variable to the system?

It is not considered as an external variable because the room of temperature 250 C is a common condition for both setup.

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Since the water for Set-up A is at room temperature and a layer of oil is on the surface of the water, no water will be lost so the mass remains the same.

For set-up B, the temperature of the water is much lower than the room temperature, therefore the water vapour in the surrounding air will lose heat and condense as water droplets onto the outer surface of the cup. This results in Set-up B becoming heavier, hence the reading on the digital balance increases.