Question

Calculate the mass of methane that must be burned to provide enough heat to convert 266.0 g of water at 39.0°C into steam at 110.0°C. (Assume that the H₂O produced in the combustion reaction is steam rather than liquid water.)

Answer 1

The amount of heat required for conversion of water to steam is ,

Q = heat change for conversion of water at 39^oC to water at 100 ^oC +heat change for conversion of water at 100 ^oC to vapour at 100 ^oC+ heat change for conversion of vapour at 100 ^oC to vapour at 11^0oC

Amount of heat released , Q = mcdt + mL + mc'dt'

                                          = m(cdt + L + c'dt')

Where

m = mass of water = 266 g

c' = Specific heat of steam = 2.1 J/g degree C

c = Specific heat of water = 4.186 J/g degree C

L = Heat of Vaporization of water = 2260 J/g

dt' = 110-100 = 10^oC

dt = 100 -39 = 61 ^oC

Plug the values we get Q = m(cdt + L + c'dt')

                                      = 674.7x10³ J

                                      = 674.7 kJ

We know that heat of combustion of methane is -674.7 kJ

Molar mass of CH₄ = 12+(4x1) = 16 g/mol

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = -890 kJ

According to the equation ,

1 mol = 16 g of methane upon combustion produces 890 kJ of heat

M g of methane upon combustion produces 674.7 kJ of heat

M = (16x674.7) / 890

   = 12.13 g

Therefore the mass of methane required is 12.13 g