Posted by Anonymous on Saturday, August 15, 2009 at 10:26pm.
What is the sign of heat(Q), Work(W), or Temperature for these following situations. Say if it's +,-, or 0.
1)You hit a nail with a hammer: +W, Q=0, T=0
2)You hold a nail over a Bunsen burner: Q=+Q, Temp:constant, W=0
3)You compress the air in a bicycle pump by pushing down on the handle very rapidly: -W, Temp:-, Q is -.
4)You turn on a flame under a cylinder of gas, and the gas undergoes an isothermal expansion:
isothermal is constant T, +Q, W is +
5)A flame turns liquid water into steam:
6)High pressure steam spins a turbine:
T inc so +, +Q, +W
7)Steam contacts a cold surface and condenses:
-Q, -W, Temp decreases
Can you check my work and correct me?
- thermodyn. please check work - Damon, Saturday, August 15, 2009 at 10:48pm
2. does the temp of the nail not increase?
3. When you squeeze gas it gets hot. If you do it rapidly enough there is no time for heat to escape.
4. That did the gas do work on?
5. Again I do not see any work being done.
6. The steam expands through the turbine, losing temperature and doing work
7. Again what work is being done?
- thermodyn. please check work - Count Iblis, Saturday, August 15, 2009 at 10:51pm
You have to define what the system is. In case of 1) if I take the system to be the nail, then
W = 0
Q = positive
T = Positive.
By definition, work is the change in itnernal energy due to a change in the external parameters of a system. In case of the nail, the external parameters (like volume) do not change. What happens is that work is performed by the person hammering the nail, but this converted to heat via friction and this heat is delivered to the nail.
The first law for the nail is:
Delta U = Q - W
I.e. internal energy gain of the nail is heat delivered to the nail minus work perfomed by the nail. The latter is zero and the former is positive, so the nail gains in internal energy, which causes the temperature to rise.
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