dr2much

I'll have to go back and see what else is going into the machine in addition to the fuel, product A. I'm certain that product B is coming out of the machine. Once I know what else is going in I will post that here in this thread.

studiot and imatfaal thank you both for your replies and your patience.

studiot thank you. I do understand that burning 1 kg of a substance results in much more than 1 kg of reaction products due to the reaction with oxygen. As you may have seen my post of 11:58 AM above mentions exhaust gases that I left out of my energy question, but which of course are part of the mass of reaction products versus reactant A. In terms of reaction products that have a (significant) calorific value product B is the only worth mentioning. It is correct that 1 kg product A put into this machine as a fuel gives 1 kg of product B (plus the exhaust gases as mentioned above). So, confirming that I do understand the role of mass and that I have factored that in properly, I come back to my equation: 20 MJ ---> 15 MJ + 10 MJ using the calorific values of product A and product B as the only measures of energy in and out (apart from the 10 MJ electricity) Can the difference be accounted for by enthalpy changes? The concept of enthalpy is one that many laymen like me struggle with, so you're replies/views will be appreciated.

I understand that there is no way to change the calorific value of a given substance. However, is calorific value enough to define the amount of energy that can be obtained from that substance. My guess is: yes if you look only at heat generated by burning the substance. What about (non-heat) energy that can be extracted from the change of the substance from a solid state into the gaseous reaction products of that substance and oxygen? If I'm not mistaken that is enthalpy change. Would that be able to balance my energy equation?

Can There is combustion exhaust gas from the machine that, after heat exchange, is released to atmosphere. Although the residual energy in that gas is minimal it will add to the energy output. In view of the expected low energy content I have not included it in my energy equation.

@ studiot and imatfaal Yes, the line: 1 kg of product A yields 1 kg of product B and C kWh of electricity. is important as to the mass argument. My confusion remains the energy. Maybe I should rephrase the questions as follows: Can a system have more energy output than input when using mainly calorific values as the measure of energy? What should I factor in, to obtain a balance in the energy input and output?

Studiot, thank you for your reply. However, doesn't the calorific value, being the heat released when a substance is combusted, already take the contribution of the reaction with oxygen (including the energy released when breaking the chemical bond of the oxygen atoms in an oxygen molecule) into account? Therefore, my question remains whether based on the calorific values and in this case electric energy is it possible that the energy output of this machine seems higher than the energy input? How to account for the positive difference?

Yes, fuel A is burned. However, I always understood that oxygen as such has no energy content (calorific value).

Let's assume a machine with substance A as the fuel. The output of the machine is electricity plus a product B. 1 kg of product A yields 1 kg of product B and C kWh of electricity. If the calorific value of product A is for example 20 MJ/kg and of product B it is 15 MJ/kg The electric output C of the machine from 1 kg of fuel A is 2,777 KWh. 1 kg A ---> 1 kg B + 2,777 kWh electricity A simple attempt to do an energy balance based on the calorific values gives the following: 20 MJ ----> 15 MJ + 2,7777 kWh 20 MJ ----> 15 MJ + 10 MJ Is this a possible situation? If so, how can it be accounted for that the energy content of the reaction product (based on its calorific value) plus the electric power produced is higher than the energy content of the fuel? Can this be explained by enthalpy changes for example vaporization and expansion of the fuel during combustion? Can someone please explain this in laymen's terms? (assuming that a machine with the input and output as described above is theoretically possible)