Question

First Law of Thermodynamics Help. Will post question below.

Answer 1

I just need someone to check my work to see if I am right. =) The internal energy of the gas in a gasoline engine's cylinder decreases by 195J. If 52.0J of work is done by the gas, how much energy is transferred as heat? Is this energy added or removed from the gas.

So the knowns are:
W=52 Joules
Delta Q = -195 Joules
Q= ?

The Equation: \[\Delta U = Q - W\]\[\Delta U -Q = Q - Q -W\]\[\Delta U -Q=W\]\[\Delta U - \Delta U - W = -\Delta U - W \]\[Q=\Delta U +W\] So this would be the equation I would use.
\[Q=-195+52\]\[Q=-143\]

The energy is removed from the gas. @osprey

Answer 2

@IrishBoy123

Answer 3

is that a piano keyboard?
hi
third line down looks iffy -W ?
Rest looks OK.
Heat goes out of gas to cold reservoir (the outside world) and external work energy so to speak - you move.

Answer 4

Thank you! And yes it is. @OpenStudyRocks5* made it for me. =)

Answer 5

Lol

Answer 6

In that case ... play on Maestro !

Answer 7

There's an algebraic error in 3rd line no doubt, but the numerical answer looks correct, so I think you might be "mis-using letters",....., forgive my candour :- |

In more descriptive and intuitive language, we can say that

\(\Delta E_{int} = Q_{in} + W_{on}\)

So if you heat up a gas and compress it, it gains energy. Intuitive. And that's all there is to know at this stage.

So:


\(-195J = Q_{in} + (\color{red}{- 52J})\)



If the gas does work, it has to have expanded. Because, for Work to have occurred, there has to be a Force acting over a Displacement. Labouring the point but here the work has been done by, and not on, the gas.

So it all connects up logically and we get the result that

\(-195J - (- 52J)= Q_{in} \)


\(-195J + 52J= Q_{in} \)

\( Q_{in} = -143J \)


the real point being that you can write the 1st Law out intuitively (no memory required) and then just plug and play:

\(\Delta E_{int} = Q_{in} + W_{on}\)