A gas company in Massachusetts charges $1.40 for 15.0 ft^3 of natural gas (CH4)

Question

A gas company in Massachusetts charges $1.40 for 15.0 ft^3 of natural gas (CH4) measured at 20.0 degrees C and 1.00 atm. Calculate the cost of heating 2.00 x 10^2 mL of water (enough tio make a cup of coffee or tea) from 20.0 degrees C to 100.0 degrees C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the hear is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2)(l)

Report your answer to the correct number of significant figures

Explanation / Answer

Given :

Charge for Methane = $1.40 for 15.0 ft^3.

Volume of water = 200 mL = 0.2 L = 200 g

Temperature = 20C = 293.15 K

P = 1 atm

Answer-

First, we will calculate amount of heat required to heat 200 mL water from 20C to 100C.

q = mass of water * specific heat capacity of water * change in temp

   = 200 * 4.18 * (100-20) = 66880 J = 66.88 kJ

As stated in question, 50% heat is lost. Therefore, actual heat required will be 66.88*2 = 133.76 kJ

We will calculate the moles of methane in 15.0 ft^3

1 Cubic Ft = 28.31 L

15 cubic feet = 141.584 L

Now we will use Ideal Gas Law,

n = P*V/R*T = 1*141.584/0.0821 * 293.15 K

n = 5.882 moles.

Heat of Combustion of Methane = 889 kJ/mol

But we require only 133.76 kJ.

Moles required for heating water = 133.76/889 = 0.15 moles of Methane.

Again using, Ideal gas law.

V = nRT/P = 0.15 * 0.0821 * 293.15 / 1 = 3.61 L = 0.1274 ft^3

$ required for Methane = 0.1274/15 *1.40 = $ 0.0118 = $ 0.012 = 12 cents

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