Concept of qsys=-qsurr

[Stella Nguyen 1J]

Hi everyone!

I was wondering if you guys could explain the concept of qsys=-qsurr and how we could recognize when to apply this concept to a problem that is given?

Thank you!

[Gillian Gabrielsen2F]

You can use this if it says "there's no heat gained/lost", thermal equilibrium, or if the system is isothermal. This is because when qsurr = -qsys, that means qsurr + qsys = 0 and that there's no net heat transfer. Since there's no net heat transfer, there isn't a change in temperature. hope this helps

[HannahRobinson3L]

Hello! Just to add on, assuming that this is a perfect system, with nothing else acting on it, all of the heat lost by the system is gained by the environment and vice versa. Hope this helps.

[Anirudh Mahadev 1G]

To add on, this can mainly be attributed to reversible systems. Since all energy lost as work is replaced by heat flowing into the system, the heat lost by the surroundings is equal to the heat gained by the system. Of course this is a purely ideal process and does not occur in real scenarios.

[Sabrina Galvan 3J]

You can use this if it says "there's no heat gained/lost", thermal equilibrium, or if the system is isothermal. This is because when qsurr = -qsys, that means qsurr + qsys = 0 and that there's no net heat transfer. Since there's no net heat transfer, there isn't a change in temperature. hope this helps

I agree that there is not a net heat transfer, since the heat lost by the system is gained by the surroundings, but there should be a change in temperature. This is given by the equation q=mCdeltaT, there the change in temperature is dependent on whether something is gaining or losing heat. In this scenario, the surroundings are gaining heat, and therefore will increase in temperature, and the system is looking heat, therefore decreasing in temperature. They will then reach thermal equilibrium after the heat transfer is finished, but this does not imply that there isn't a change in temperature.

[Edgar Velazquez 2K]

You can think of it as the heat that is lost (or gained) by the system is the same heat that is gained (or lost) by the surroundings, but the sign is opposite because of the direction that the heat is flowing. In equation form, you can think of it as q sys (heat of system) + q surr (heat of the surroundings) = 0. Then you can do the math (subtracting one of the terms to the other side) to find that q sys = - q surr or q surr = -q sys. The numerical value of the heat is the same for both, but the sign is opposite since what is being lost by one is gained by the other.

[Sofia Azmat 1I]

Think of it as heat gained vs heat lost. When heat is lost by the system it is GAINED by the surroundings, almost like an inverse reaction. When heat is lost by the surroundings it is GAINED by the system. This is of course assuming the system has no outside factors to account for.

[Jarrett Sung 3B]

Think of qsys = -qsurr as the heat lost by the system is gained by the surroundings or vice versa. This is assuming a perfect system, but unless otherwise stated I think we should assume a perfect system.