A puddle holds 150 g of water. If 0.5 g of water evaporates from the surface what is the approximate temperature change of the remaining water?
The puddle has 150 g of water. 0.5 g of the water evaporates. The heat of vaporization of water is 2260 kJ/kg or 2.260 kJ/g or 2260 J/g. That is equivalent to saying that each gram of water needs 2260 J of energy to convert from a liquid to a gaseous state.
When 0.5 g of water evaporates it consumes 1130 J of energy.
The specific heat of water is 4.186 J/g. A 1 degree drop of temperature in 1 gram of water requires a reduction of 4.186 J of energy.
Now when 0.5 g of water evaporates, we are left with 149.5 g of water. And the reduction in energy is by 1130 J.
This means a reduction in temperature of 1130/ (149.5*4.186) degree or 1.805 degree Celsius.
The required drop in the temperature of the water left after 0.5 g has evaporated is 1.805 degree Celsius.
We assume that the latent heat of vaporisation for water is L (2260j/g).
So for the evaporation of 0.5 g of water, the loss of heat energy from the water is 0.5 *L = 2260*0.5 = 1130 J = 1130/4.2 calories = 269.05 calories.
The remaining water after evaporation of 0.5 gram of water = (150-0.5) g = 149.5.
Therefore the reduction in temperature of the remaining water after evaporation of 0.5 gram of water = Heat loss in calories/mass of remaining water = 269.05/149.5 = 1.7996 Celsius = 1.8 degree C approximately.