What is the Maximum Temperature for a Copper Block at 1000 atm Pressure?

AI Thread Summary
The discussion revolves around calculating the final pressure of a copper block when its temperature is increased from 5°C to 10°C at constant volume, using the formula that relates pressure, volume expansivity, and isothermal compressibility. The calculated final pressure for part (a) is 40200 Pa, which is significantly lower than the maximum pressure of 1000 atm. For part (b), the initial assumption leads to an incorrect final temperature of 2.4°C, while the expected answer is 17°C. The calculations indicate that the values for volume expansivity and isothermal compressibility must be carefully applied to determine the maximum temperature the system can withstand. The discussion highlights the importance of accurate calculations and understanding the relationships between pressure, temperature, and material properties.
Muthumanimaran
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Homework Statement


(a) A block of copper at a pressure of 1 atm and a temperature of 5 degree celcuis is kept at constant volume. If the temperature is raised to 10 degree celcius what will be the final pressure?

(b) If the vessel holding the block of copper has a negligibly small thermal expansivity and can withstand a maximum pressure of 1000atm, what is the highest temperature to which the system may be raised?

(Note: The volume expansivity and isothermal compressiblity are not always listed in handbooks of data. However, volume expansivity is three times the linear expansion coefficient and isothermal compressibility is reciprocal of the bulk modulus. For this problem, assume that volume expansivity and isothermal compressibility remain practically constant within the temperature range of 0 to 20 degree celcius at the values of 4.95 x 10^(-5) K^-1 and 6.17 x 10^(-12))

Homework Equations


(Pf-Pi)=(volume expansivity / isothermal compressibility) (Tf-Ti)

The Attempt at a Solution


part (a) of the problem can be easily done by using by substituting the given data in the formula above.
$$P_{f}-P_{i}=\frac{\beta}{\kappa}(T_{f}-T_{i})$$
$$P_{f}-10^{5}=\frac{4.95 \times 10^{-5}}{6.17 \times 10^{-12}} (10-5)$$
gives Pf=40200 Pa

If I do the same for part (b) by taking initial temperature as 0 degree celcius and initial pressure as 1 atm, the final temperature is 2.4 degree celcius but the correct answer is 17 celcius, give me a hint.
 
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What answer to you get for part (a)? Also, please show your calculations.
 
DrClaude said:
What answer to you get for part (a)? Also, please show your calculations.
$$P_{f}-P_{i}=\frac{\beta}{\kappa}(T_{f}-T_{i})$$
$$P_{f}-10^{5}=\frac{4.95 \times 10^{-5}}{6.17 \times 10^{-12}} (10-5)$$
gives Pf=40200 Pa
 
Muthumanimaran said:
Pf=40200 Pa
You are off by a few orders of magnitude.

In any case, your result for Tf = 10 °C is Pf < 1000 atm. How can you get a temperature of 2.4 °C for 1000 atm?
 
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