How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from 95.0°C to 45.0°C? Assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 kcal/g). Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answe

31.3 g

The answer is higher than the true answer.

Explanation:

By neglecting the heat lost by other processes, the energy conservation states that:

Qcooling + Qevaporate = 0

The cooling process happens without phase change, so the heat can be calculated by:

Qcooling = m*c*ΔT

Where m is the mass, c is the heat capacity (cwater = 4184 J/kg. K), and ΔT is the temperature variation (final - initial).

The evaporate process happen without changing of temperature (pure substance), and the heat can be calculated by:

Qevaporate = m*L

Where m is the mass evaporated and L is the heat of evaporation (2340000 J/kg).

0.350*4184 * (45 - 95) + m*2340000 = 0

2340000m = 73220

m = 0.0313 kg

m = 31.3 g

Because of the assumptions made, the real mass is not that was calculated. There'll be changing mass when the coffee is cooling, and there'll be heat loses by other processes because the system is not isolated. Also, the substance is not pure. So, there'll be more factors at the energy equation, thus, the answer is higher than the true answer.