I took a quiz, and wanted to see if I got any of the following problems right or
ID: 2994471 • Letter: I
Question
I took a quiz, and wanted to see if I got any of the following problems right or wrong.
2. The Helmolz function is a combination property defined by the sum of internal enegery plus the negative value of temperature times entropy. Consider a closed system that undergoes a cycle which starts with the initial pressure equal to 600 kPa and initial Helmholz function equal to 300 kJ/kg. During the cycle 200 kJ of heat was transferred to the system. What is the value of the Helmholz function after the cycle is complete?
---- I figured since it was a closed system, the final function equals initial function, so I put 300 kJ/kg as my answer for this one.
3. A refrigerator must transfer 1000 btu/hr to the surrounding at 68 degrees fahrenheit to maintain its refrigerated space at 20 degree fahrenheit. The refrigerator requires 200 btu/hr to accomplish this. What is the maximum possiple coefficient of performance?
---- I used COP = 1 / ((T,H / T,L) - 1) to solve this (with converting the temperatures to Rankine), and got 9.99313 (or about 10)
5. A 1 kg block of copper with specific heat 0.386 kJ/kg K is initially at 300 K. It is then put in contact with a thermal reservoir in the surroundings at 500 K until it reaches thermal equilibrium with it. (b) What is the change of entropy of the surroundings? (c) How much entropy was generated during the process?
---- For part B, I used S = Cp x ln (T,2 / T,1) and got 0.514 kJ/K
---- For part C, I used S,gen = S,sys + S,surr and got 0.711 kJ/k
*** If any parts are incorrect then please show the correct steps for solving the problems ****
*** I will only award points to the people who show me proof that my answers are either correct or incorrect ***
Explanation / Answer
We cannot predict because we do not know whether the process was constant pressure, constant volume, constant temperature, adiabatic etc.???
Cooling done = Heat rejected - work input
= 1000 - 200
= 800 Btu / hr
COP = Cooling done / Work input
= 800 / 200
= 4
5. You have done it correctly.