Chemistry - Thermochemistry Question with Solution | TestHub
Assume that for a domestic hot water supply, 160 kg of water per day must be heated from 10°C to 60°C and gaseous fuel propane, C3H8, is used for this purpose. What volume of propane gas at STP would have to be used for heating domestic water, with efficiency of 40%? Heat of combustion of propane is -500 kcal/mol and specific heat capacity of water is 1.0 cal/K-g.
Options:
Answer:
Solution:
The heat required is calculated as:
Q = mcΔ T
Given m = 160000 g, c = 1.0 cal/g°C, and Δ T = 50°C.
Q = 160000 g × 1.0 cal/g°C × 50°C = 8.0 × 106 cal = 8000 kcal
With an efficiency of 40%, the total heat that must be supplied is:
Qsupplied = 8000 kcal/0.40 = 20000 kcal
The number of moles of propane needed, given that 500 kcal are produced per mole, is:
Moles of propane = 20000 kcal/(500 kcal/mol) = 40 mol
At STP (273 K, 1 bar), the molar volume is 22.7 L/mol. Therefore, the volume of propane required is:
V = 40 mol × 22.7 L/mol = 908 L
This corresponds to option (b).
