Question

John Smith is a chemistry student who lives in Florida and the mosquitoes are horrible this year. He is sweating…thinking…why don’t I live in Colorado, I would be much cooler. He is thinking about phase changes, intermolecular forces and the many natural phenomena that occur around him every day.

a) He has a glass of cold lemonade with ice in it and he notices that his glass is also sweating. What phase change was occurring when sweat was forming on the outside of his glass and why(how) was it happening?

What about the ice in his drink. It was definitely melting. How much energy does it take to convert ice to steam anyway, which is essentially what he was doing by drinking, then sweating

Calculate the energy required to convert John’s 10g ice cube, which began at -3^oC, to water vapor at 100^oC. Show all your work.

Note:ice began at -3 degrees celcius shown above

Answer 1

1) When ice is kept in the glass, the temperature of glass surface would become lower. The glass seems to sweating, that is the atmospheric water vapor are getting condensed on the chilled wall of the glass. When water vapors are condensed, it gets converted into liquid. That is why he is observing sweat (condensation of atmospheric water vapor) on the wall of the glass. Hence, in this case, gaseous phase is converted to liquid phase.

(2)

We know that
Specific heat capacity of water = 2.1 J g^-1oC^-1

Specific heat capacity of water = 4.2 J g^-1oC^-1
Specific heat capacity of water = 4.2 J g^-1oC^-1
Latent heat of vaporization of water = 2256 J g^-1
Latent heat of fusion (melting) of ice = 336 J g^-1

Heat to get ice from – 3^oC to 0^oC
Q1= (10 g) (2.1 J g^-1oC^-1) (0^oC-(-3^oC)) = 63 J
Heat to change ice at 0^oC to liquid water at 0^oC

Q2 = (mass) (Latent heat of fusion of water)
= (10 g) (336 J g^-1) = 3360 J

Q3 = heat needed to get liquid water from 0^oC to 100 C
= (mass) (C_WATER) (100^oC – 0^oC)
= (10 g) (4.2 J g^-1oC^-1) (100^oC) = 4200 J

Heat to change water at 100^oC to water vapor at 100^oC
Q4 = (mass) (Latent heat of vaporization of water)
= (10 g) (2256 J g^-1) = 22560 J

Total heat = Q1 + Q2 + Q3 + Q4
= (63 J) + (3360 J) + (4200 J) + (22560 J)
= 30183 J = 30.18 kJ