Thermochemistry: Heating water with Methane

Question

I'm stuck on this one

 

How many grams of methane must be combusted to heat 1.00 kg of water from 25.0 °C to 90 °C, assuming H₂O(l) is the product of the reaction and no heat is lost to the surroundings?

 

So far I have 

1000 g H2O

ΔT = 65.0 °C

q = 0

 

CH₄ (g) + 2O₂ (g) → 2H₂O (l) + CO₂

 

heat evolved = 4.1798 J g^-1 °C^-1 (1000g)(65.0°C)(1kJ/1000J) = 271.6 kJ

Answer 1

Hi Alex,

 

From the equation above (CH₄ (g) + 2O₂ (g) → 2H₂O (l) + CO₂),  Δ H = -802.4 kJ, if we show water vapor rather than liquid water as a product.

 

Reversing  the melting of ice and the combustion of methane, the thermochemical equation

(2H₂O (l) + CO₂ → CH₄ (g) + 2O₂ (g) )   Δ H = 890.4 kJ

and what was an endothermic process becomes exothermic, and vice versa.

2H₂O (s) →  2H₂O (l)     Δ H = 2(6.01 kJ) = 12.0 kJ

 

 

the enthalpy change is -802.4 kJ rather than -890.4 kJ because 88.0 kJ are needed to convert 2 moles of liquid water to vapor

2H₂O (l) →  2H₂O (g)     Δ H = 88.0 kJ

 

Remember this? Δ H = H (products) - H (reactants)

 

CH₄ (g) + 2O₂ (g) → 2H₂O (l) + CO₂, Δ H = -802.4 kJ

 

With these being said, where did you get 4.1798 J /g C?

I'm asking you this because specific heat of water is 4.184 J /g C.

 

You must find this from your Chemistry Textbook to do this

Δ H_f (kJ /mol)   standard enthalpies of Formation of Inorganic Substance at 25 C.

 

By the way, molar mass of methane (CH₄) is 16.01 g /mol.  And there are 2 mol of H₂O in this equation.  And molar mass of water is 18.0 g /mol.  You might need these so you can do the stoichiometry to solve this problem.

 

Good luck to you.