Water has a specific heat of 4.18 j/gk and a molar heat of vaporization of 40.7 kj. calculate the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c

To calculate the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c

Explanation:

We will use heat balance equation

Q = q1+q2 = mCΔt + nL

where

m = mass of water = 90g

C = specifi heat of water =4.18J/gK

L = Latent heat of Vaporization = 40.7kJ/mol

Δt = temperature changes = 100 - 20 = 80 degree

water at 20 degrees to water at 100 degrees

then water at 100 degrees to vapor at 100 degrees

q1 = (90g)(4.18J/gK)(80K)

q1 = 30096J=30.1kJ

But

m =90 g water, gives  n = 5mol water (divide by his molar mass which is 18g)

q2=nL= (40.7kJ/mol)(5mol)

q2=203.5kJ

the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c = q1+q2 =30.1+203.5=233.6kJ

The dry bulb temperature is the ambient air temperature that is measured by regular thermometers, while the wet bulb temperature is measured by thermometers that are wrapped in  wicks. The greater the wet bulb depression, the greater the felt effect is on the discharge air temperature.

Explanation: