Question

If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of water can be heated from 20.0°C to 100.0°C? (Assume that the density of water is 1.00 g/mL)

Answer 1

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

Combustion:

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

$Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J$

Calorimetry:

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

$Q=mC\Delta T$

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

$m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g$

Finally, we convert it to liters as required:

$V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL} \\\\V=42.27L$

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