The total energy released per gram of methane in the experiment is 119941.33 J/g
How to determine the change in the temperature of water
- Initial temperature of water (T₁) = 25 °C
- Final temperature of water (T₂) = 68 °C
- Change in temperature (ΔT) = ?
Change in temperature (ΔT) = T₂ – T₁
Change in temperature (ΔT) = 68 – 25
Change in temperature (ΔT) = 43 °C
How to determine the heat absorbed by the water
The absorbed by the water can be obtained as illustrated below:
- Mass of water (M) = s00 g
- Change in temperature (ΔT) = 43 °C
- Specific heat capacity of the water (C) = 4.184 J/gºC
- Heat (Q) =?
Q = MCΔT
Q = 500 × 4.184 × 43
Q = 89956 J
How to determine the energy released by methane in the experiment
- Heat absorbed by water = 89956 J
- Percentage of heat absorbed by water = 30%
- Heat released by methane =?
Heat absorbed by water = 30% of heat released by methane
89956 = 30% × heat released by methane
89956 = 0.3 × heat released by methane
Divide both sides by 0.3
Heat released by methane = 89956 / 0.3
Heat released by methane = 299853.33 J
How to determine the heat released per gram of methane
- Heat released by methane (Q) = 299853.33 J
- Mass of methane (m) = 2.5 g
- Heat per gram (ΔH) =?
Q = m × ΔH
Divide both sides by m
ΔH = Q / m
ΔH = 299853.33 / 2.5
ΔH =119941.33 J/g
Learn more about heat transfer:
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