Question

Calculate the mass of methane that must be burned to provide enough heat to convert 140.0 g of water at 19.0°C into steam at 115.0°C. (Assume that the H2O produced in the combustion reaction is steam rather than liquid water.)

Answer 1

The amount of heat required for the conversion of water at 19.0°C into steam at 115.0°C is ,

Q = heat change for conversion of water at 19.0^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 115 ^oC

Q = mcdt + mL + mc'dt'

   = m (cdt + L+ c'dt')

Where

m = mass of water = 140.0 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 = 100-19.0 = 81.0 ^oC

dt' = 115.0 - 100 ^oC = 15.0 ^oC

Plug the values we get

Q = 140.0 x [(4.186x81.0) + 2260 + (2.1x15.0 ) ]

    = 368279 J

    = 368.279 kJ

The combustion of methane is   CH₄(g) + 2O₂ (g) CO₂(g) + 2H₂O(l)   ; H = ?

ΔH = ΔH⁰_fproducts – ΔH⁰_freactants

ΔH = [(ΔH⁰_fCO2(g) )+ (2 x ΔH⁰_{fH2O (l)} )]- [( ΔH⁰_fCH4(l) )+(2 x ΔH⁰_{fO2 (g)} )]

      = [(-393.51)+(2x(-285.83))] - [(-74.85)+(2x0)]

     = -890.32 kJ

CH₄(g) + 2O₂ (g) CO₂(g) + 2H₂O(l)   ; H = -890.32 kJ

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

According to the balanced equation,

1 mole of CH₄ produces 890.32 kJ

                      OR

16 g of CH₄ produces 890.32 kJ

M g of CH₄ produces 368.279 kJ

M = (16x368.279) / 890.32

    = 6.62 g

Therefore the mass of methane that must be burned is 6.62 g