Question

The heat of combustion of propane, C3H8, is 2220 kJ/mol. The specific heat of water is 4.184 J/g·°C. Mrs. S lives in the country and uses propane to heat things. How many litresof propane at STP must be burned to raise the temperature of 1.00 L of water from 20.0°C to 100.0°C so that Mrs. S can make herself a nice cup of tea? Assume that no heat is lost to the surroundings and remember that the density of water is 1.00 g/mL.And if propane costs 88.9 cents per litre, how much did the propane cost to heat the water?

Answer 1

(a) 3.38 Litres (b) $3‬.00

Step-by-step explanation:

Q=Quantity of Heat required to make the cup of tea

Q = mcΔФ

m = mass of water

c= Specific capacity of water, 4.184 J/g/° C

p = density of water, 1g/mL

v = volume of water for tea, v= 1L or 1000mL

ΔФ= Change in Temperature , T₂ -T₁, T₂= 100°C, T₁=20°C

m = 1 g/mL x 1000 mL

= 1000g

Q= 1000 x 4.184 (100-20)

= 1000 x 4.184 x 80

= 334,720‬J

Number of moles of propane required = Q/ heat of combustion

= 334,720‬/ 2220000

= 0.150774 moles

At STP, 1 mole of every gas occupies 22.4 dm³

0.150774 moles of propane will occupy = 0.150774 x 22.4

= 3.3773376 ‬dm³

= 3.3773376 Litres

≈ 3.38 Litres

If the cost per litre of propane is 88.9cent or $0.889

The total cost of propane for the tea will be = 0.889 x 3.38

= $3.00482

≈ $3‬