Answer to Question #200418 in Physics for M.m

Question

We have 2 kg of ethanol at a temperature of 30 degrees Celsius. The boiling point of ethanol is 80 degrees Celsius. If the specific heat of ethanol is 2400 kJ / kg Kelvin and the latent heat of vaporization is 850000 kJ / kg, how much heat is needed to evaporate 200 g of this ethanol at 80 ° C?

Answer 1

$$Q=mc\Delta T+mR\\Q=(0.2)(2400)(80-30)+(0.2)(850000)\\Q=194\ kJ$$