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Yes.
It seems our formulas are different. I mean the structure of them. But either ways they give the same answer.
It seems our formulas are different. I mean the structure of them. But either ways they give the same answer.
No, it's fine, I'm just trying to get you to be clearer and more careful in writing out your work, as much for your benefit as mine.Can we please do part C?. I know this is very bad and irritating for you. I am sorry for the trouble.
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.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.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.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.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.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.the volume of the steel?
The 2cm is the internal diameter, not the external diameter.(π(0.01)^2 - π(0.009)^2)*
No, the gas is cooling from the high temperature found previously to T_{f}. 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.MgCg(Tf- 18)
whoops! That's wrong.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.Excuse me, Mr.Haruspex it's getting quite late for me could you please show me what you mean?.