Question

Natural gas (CH₄) is burned to raise the temperature of 20 L of water in a storage tank from 5.8 °C to 82.0 °C.

a) How much heat energy (in J) is absorbed by the water?

b) Assuming this was a closed system, how much heat energy (in J) was released by the burning natural gas?

c) What mass (in kg) of methane was burned?

d) If some heat released by the burning of natural gas was “lost” heating up the tank that holds the water, would this make the actual methane mass consumed higher or lower than what you just calculated? Why?

Answer 1

volume of water= 20 L =20*1000 ml, density of water ( assumed)= 1 g/ml

mass of water= Volume* density= 20*1000*1= 20000 gm

specific heat of water = 4.184 J/gm.deg.c

heat added to water due to combustion = mass of water* specific heat of water* temperature rise= 20000*4.184*(82-5.8) joules =6376416 joules

this much energy has come from burning of natural gas ( since the system is closed, all the heat is taken by water)

heat of combustion of methane= -802.3 Kj/mole

1 mole of methane gives =802.3*1000 Joules of heat

hence moles of methane burned= 6376416/802.3*1000= 7.95 moles

mass of methane= moles* molar mass= 7.95*16= 127.2 gm=127.2/1000 kg=0.1272Kg

when some of the heat is lost, more methane is needed to rise the temperature of water to account for the heat loss as well as the temperature rise of 20 L of water from 5.8 to 82 deg.c.