Thermdynamics

[rsd_sosu]

Would Someone check my work please?
Here is the question.
The pressure on 500g of copper is increased reversibly and isothermally from 0 to 5000atm.
Where
density = 8.96X10^3kg/m^3
Volume Expansivity = 49.5X10^-6K^-1
isothermal compressibility = 6.18X10^-12Pa^-1
specific heat = 385 J/Kg*K
temp = 298K

Now the first part of the question ask how much heat is transferred during the compression? (I worked this out but got a different answer than the book )anyway here it is.
Q = -T*V*(Volume Expansivity)*(P_f-P_i)
Q = -(298K)*(5.58X10^-5m^3)*(49.5X10^-6K^-1)*(5.056X10^8Pa)
Where i found Volume From Mass/density
and changed Pressure from atm to Pascals
after pluging in all these numbers my answer is -416 N*m or -416J
the answer the book gives is -139.9J

Can anyone see the problem here?
sorry if there is any confusion with symbols this is my first post and do not know how to use the code yet.
Thanks

[Bystander]

"Isothermal process." Ask yourself what is happening in an "isothermal process."

[rsd_sosu]

Ok I asked my self and I replied the temperature is holding at 298K. What r u getting at? In order for the temp to stay the same heat must leave during compression. Precisely how much is the question being asked.

[Doc Al]

Originally posted by rsd_sosu
Now the first part of the question ask how much heat is transferred during the compression?
How much work is done on the copper as it is compressed?

[rsd_sosu]

I talked to my Professor today and he informed me that the answer in the book was indeed incorrect. This is not the first time and probably won't be the last error i find. the text i am using is Heat And Thermodynamics by Zemansky and Dittman which is full of errors in the seventh edition. Anyway thanks for the responses.

The Work involved that you (Doc Al) mentioned was in fact the second of four questions asked. Here it is.


W = \int_{P_i}^{P_f} VkP dP = \frac{1}{2} V k P^2_f




\frac{1}{2}(5.58X10^{-5}m^3)(6.18X10^-^1^2Pa^{-1})(5.07X10^8Pa)^2




= 44.3N-m =44.3J


Just to get a bit more practice with this scrip I will show the Change in internal energy as well, which was the third question.


\Delta U = Q + W = -417J + 44.3J = -372.7J


Later,