Question

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 answer?

a) 82.0 g; Lower
b) 115.0 g; Higher
c) 152.5 g; Lower
d) 195.0 g; Higher

Answer 1

To cool the coffee and the cup from 95.0°C to 45.0°C, approximately 115 grams of coffee must evaporate.

Step-by-step explanation:

To cool the coffee and the cup from 95.0°C to 45.0°C, we need to calculate the amount of coffee that needs to evaporate. The heat required to cool the coffee is given by the formula: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of water is 4.18 J/g°C, and the change in temperature is 50.0°C. The heat required is Q = (350 g + 100 g) * 4.18 J/g°C * 50.0°C = 87650 J.

The heat of vaporization is 2340 kJ/kg (560 cal/g), which is equivalent to 2340000 J/kg (560000 cal/g). Let's assume x grams of coffee evaporates. The heat required for evaporation is Q = x g * 2340000 J/kg. Equating the two expressions for Q, we get:

(350 g + 100 g) * 4.18 J/g°C * 50.0°C = x g * 2340000 J/kg

Simplifying the equation, we find that x = 115 g.