For getting the answer to Part C, we need to solve Part A and Part B (a) Since t
ID: 700688 • Letter: F
Question
For getting the answer to Part C, we need to solve Part A and Part B
(a) Since the container is rigid and strong, hence the change in volume Delta v = 0, which will imply the work done will be zero
W = 0
Since we are heating, hence q must be positive in this case, now we now the formula
Delta U = q + w, since Delta w is zero which means Delta U must also be positive
(b) Since the container is rigid and strong, hence the change in volume Delta V = 0, which will imply the work done will be zero
Delta W = 0
Since we are cooling , hence Delta q must be negative in this case, now we now the formula
Delta U = Delta q + Delta w, since Delta w is zero which means Delta U must also be negative
c)
Now since q1 is positive and q2 is negative, hence the sign of (q1+q2) can be positive or negative depending upon which quantity is higher in magnitude
Threfore there is no information that how much amount of heat spent is there in cooling and heating cycle and there are more variables which will require more information to predict the sign of q
Explanation / Answer
Please do read the full post before starting to answer my question
Consider the following question:
In part C, the answer key says that q1 + q2 can be positive, zero, or negative.
Could please explain why the system may need to lose more/less thermal energy to cool back to its original temperature?
9. A chemical system is sealed in a strong, rigid container at room temperature, and then heated vigorously (a) State whether A, q, and w of the system are positive, negative, or zero during the heating process. (b) Next, the container is cooled to its original tempera- ture. Determine the signs of U,q, and w for the cool. ing process. mine the signs of AU, + U2), (J1 + q2), and (w, + IV), if possible.Related Questions
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