Yes, but we've done that part. We are now interested in what happens as heat from the compressed gas is transferred to the steel.Akaramos45 said:Doesn't the gas gain energy during the compression due to the work done?
Ok. I would have said it a bit differently... The specific heat of the gas is very small compared to that of the cylinder, so nearly all the gained internal energy will finish up in the steel.Akaramos45 said:The teacher also said I could assume that the final temperature is close to the initial temperature, so that all the of the energy in the hot gas is transferred to the steel
Yes. The steel will heat up and the gas cool down until they reach the same temperature. It does not say so, but you need to ignore any heat loss from the cylinder to the surroundings.Akaramos45 said:The steel and the compressed gas come into thermal equilibrium. Does that mean due to a difference in temperature heat transfer occurs?
You can assume it started at the same temperature as the gas did, 18C.Akaramos45 said:How do I calculate the increase in the the temperature of the steel if i do not know the initial temperature of the steel
You know the heat energy going into it, its volume, density and specific heat. If you don't know an equation connecting those with temperature change, look up specific heat.Akaramos45 said:I see. How do I calculate the increase in temperature of the steel. What formulas would I use?.
I am sorry this is my first doing this sort of problem. Could you show me an example?
Yes. You are given the internal diameter, the length and the thickness.Akaramos45 said:the volume of the steel?
The 2cm is the internal diameter, not the external diameter.Akaramos45 said:(π(0.01)^2 - π(0.009)^2)*
No, the gas is cooling from the high temperature found previously to Tf. But anyway, we established that you can just write that the heat transferred to the steel is the whole of the internal energy gained by the gas during compression, and you already calculated that.Akaramos45 said:MgCg(Tf- 18)
whoops! That's wrong.haruspex said:But anyway, we established that you can just write that the heat transferred to the steel is the whole of the internal energy gained by the gas during compression, and you already calculated that.
Sorry, but I'm not that expert in this area myself, and I don't want to lead you astray. I need to check some things before I give a firm answer. I understand that this will be too late for you tonight.Akaramos45 said:Excuse me, Mr.Haruspex it's getting quite late for me could you please show me what you mean?.
Will all of the gained internal energy will be transferred? Even when the gas cools back to 18C it will be at a higher pressure than it started. That represents a gain in energy, no?Akaramos45 said:I was thinking since we can assume that all of the energy of the hot gas is transferred to the steel. Wouldn't what you said before still work?
I would not have thought so, but I could be quite wrong about this. Maybe the internal energy is just PV, so it all does end up in the steel.Akaramos45 said:Could i say that the increase in energy due to an increase in pressure is neglible?
In my judgment, the way he did it in the first place is correct. The assumption is that the gas adiabatically heats up first, and then equilibrates thermally with the cylinder. So, the way you originally had it doped out was correct.haruspex said:I would not have thought so, but I could be quite wrong about this. Maybe the internal energy is just PV, so it all does end up in the steel.
@Chestermiller , you'll be able to settle this immediately, I'm sure.
Yes, I understand that, but I had this nagging thought that surely the compressed gas, even after equilibration, is storing energy. But I now think that its potential to do work comes from its low entropy, in the same way that an uneven temperature distribution has the capacity to do work.Chestermiller said:the gas adiabatically heats up first, and then equilibrates thermally with the cylinder.
Rather than assuming the the gas heats up first and then equilibrates with the cylinder, it is also possible to assume that the two heat up simultaneously. In this case $$dU=-PdV$$or $$MCdT=-\frac{nRT}{V}dV$$or equivalently $$MC\frac{dT}{T}=\frac{nR}{V}dV$$where M is the mass of the cylinder and C is the heat capacity of the steel. I think that this is more likely what they had in mind. In this case, the expansion would be nearly isothermal (because of the large thermal inertia of the cylinder).haruspex said:Yes, I understand that, but I had this nagging thought that surely the compressed gas, even after equilibration, is storing energy. But I now think that its potential to do work comes from its low entropy, in the same way that an uneven temperature distribution has the capacity to do work.
Thanks for clearing it up.
The question statement says the air is compressed adiabatically.Chestermiller said:it is also possible to assume that the two heat up simultaneously... I think that this is more likely what they had in mind.