Final answer:
To heat 0.8 kilograms of water by 13.0 °C, you need to burn approximately 0.772 grams of methane.
Step-by-step explanation:
To calculate the amount of methane that must be burned to heat 0.8 kilograms of water by 13.0 °C, we need to use the equation:
1 mole of CH₄ + ΔH = 891 kJ
From the equation, we can see that 1 mole of CH4 releases 891 kJ of heat energy. To calculate the amount of methane needed, we need to convert 0.8 kilograms of water to grams:
0.8 kg x 1000 g/kg = 800 grams
Next, we need to calculate the heat required to heat the water by 13.0 °C:
q = mcΔT
q = (800 g) x (4.18 J/g °C) x (13.0 °C) = 43,232 J
To convert J to kJ, divide by 1000:
43,232 J ÷ 1000 = 43.232 kJ
Now we can calculate the amount of methane needed:
(43.232 kJ) ÷ (891 kJ/mol) = 0.048 mol
Converting mol to grams:
0.048 mol x (16.04 g/mol) = 0.772 g
Therefore, you need to burn approximately 0.772 grams of methane to heat 0.8 kilograms of water by 13.0 °C.