How much work from an engine?

[megid]

Homework Statement

An internal combustion engine operated with methanol (CH3OH) as the fuel has a thermodynamic efficiency of 20%.
a) How much work is derived from burning one mole of ethanol in the engine?
b) What is the max work that could be derived by reacting the methanol with pure oxygen in a fuel cell at 25 C?

Homework Equations

For CH3OH + (3/2)O2 -> CO2 + 2H2)
G = -702.36 kJ at 25 C
H = -726.51 kJ at 25 C

The Attempt at a Solution

Not too sure how to even start this problem, as I don't know what relevant thermodynamic equations I need to use. Hoping someone could please point me in the right direction.

 

[Andrew Mason]

Homework Statement

An internal combustion engine operated with methanol (CH3OH) as the fuel has a thermodynamic efficiency of 20%.
a) How much work is derived from burning one mole of ethanol in the engine?
b) What is the max work that could be derived by reacting the methanol with pure oxygen in a fuel cell at 25 C?

Homework Equations

For CH3OH + (3/2)O2 -> CO2 + 2H2)
G = -702.36 kJ at 25 C
H = -726.51 kJ at 25 C

The Attempt at a Solution

Not too sure how to even start this problem, as I don't know what relevant thermodynamic equations I need to use. Hoping someone could please point me in the right direction.

How much energy is released from the combustion of one mole of CH₃OH? That is the input energy. If the efficiency is 20%, how much of that is converted into work?

For part b) you have to figure out the combustion temperature of the methanol and then work out the maximum efficiency of a reversible cycle operating between that temperature and 25C (298K).

AM

 

[megid]

So the energy release is -702.36 kJ and 20% of that would be -140.47 kJ which would be the work done?

Also, I haven't really a clue how to start b even after the hint you gave me... I don't know how to find the combustion temp of methanol. I'm sorry for all the questions, but I really am lost. I don't want anyone to hold my hand, because I really want to understand this.

 

[Andrew Mason]

So the energy release is -702.36 kJ and 20% of that would be -140.47 kJ which would be the work done?

If your figures are correct, that is correct.

Also, I haven't really a clue how to start b even after the hint you gave me... I don't know how to find the combustion temp of methanol. I'm sorry for all the questions, but I really am lost. I don't want anyone to hold my hand, because I really want to understand this.

If you combust one mole of methanol, how many moles of O₂ do you consume? How many moles of CO₂ and H₂O does that combustion produce? What are the specific heats at constant volume of these products of combustion? Now add the heat energy released in that combustion and determine the resulting change in temperature: $\Delta T = \Delta U/nC_v$

AM

 

[megid]

For part b, I don't want to seem stupid, but are you sure that's the method for fuel cells? Because it seems like you're still talking about combustion engines.

 

[Andrew Mason]

For part b, I don't want to seem stupid, but are you sure that's the method for fuel cells? Because it seems like you're still talking about combustion engines.

The energy produced in the fuel cell is from the combustion of methanol. The fuel cell somehow turns that heat energy into electricity in an efficient way. But the maximum efficiency is determined by the thermodynamic limit, regardless of the technology. That is simply a matter of the second law of thermodynamics.

In looking at your previous answer, you have to use the $\Delta H$ of the reaction, not the $\Delta G$.

AM