Determining the efficiency of a water heater.
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Some of the butane, C4H10(g), in a 200.0 liter cylinder at 25.0 degrees C is withdrawn and burned at a constant pressure in an excess of air. As a result the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 liters of water from 25.0 to 62.2 degree C. Assume that the combustion products are CO2(g) and H2O(l) exclusively. Determine the efficiency of the water heater, i.e., what percent of the heat of combustion was absorbed by the water?
Heat capacity of water: 4.18 J/̊C/ g;
density of water: 1.00 g/mL
(C4H10) = -125.6 kJ/mol
(CO2) = -393.5 kJ/mol
(H2O) = -286 kJ/mol.
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Solution Summary
This solution is given in seven steps with explanation of determining the efficiency of a water heater.
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First, write and balance the combustion reaction of C4H10
C4H10 + 6.5O2 ---> 4CO2 + 5H2O
Second, calculate the enthalpy (or heat) of the reaction
Heat = [4(-393.5) + 5(-286)] - [(-125.6) + (0)] = -2878 kj/mole
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