Question

Calculate the mass of methane that must be burned to provide enough heat to convert 270.0 g of water at 24.0°C into steam at 116.0°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 24^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 116^oC

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

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

Where

m = mass of water = 270 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' = 116-100 = 16^oC

dt = 100 - 24 = 76 ^oC

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

                                      = 705.2x10³ J

                                      = 705.2 kJ

We know that heat of combustion of methane is -890 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 705.2 kJ of heat

M = (16x705.2) / 890

   = 12.68 g

Therefore the mass of methane required is 12.68 g