ThermochemistryHard
Question
Assume that for a domestic hot water supply, 160 kg of water per day must be heated from 10oC to 60oC 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
A.896 L
B.908 L
C.896 m3
D.908 m3
Solution
Heat liberated from propane = Heat absorbed by water
Or, $n \times 500 \times 10^{3} \times \frac{40}{100} = 160 \times 10^{3} \times 1 \times 50 \Rightarrow n = 40$
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