In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energy decreases by 4.07 x 104 J.
Determine each of the following quantities (including the algebraic sign):
(a)W,
(b)U
(c)Q.
The Correct Answer and Explanation is :
To solve this problem, we’ll use the first law of thermodynamics, which states that:
[
\Delta U = Q – W
]
Where:
- ( \Delta U ) is the change in internal energy.
- ( Q ) is the heat transferred to or from the system.
- ( W ) is the work done by or on the system.
Given:
- The work done by the student, ( W = 1.82 \times 10^4 \, \text{J} ).
- The change in internal energy, ( \Delta U = -4.07 \times 10^4 \, \text{J} ) (since it decreases, it’s negative).
Step-by-step calculation:
(a) Work (W):
The work done is already given in the problem:
[
W = 1.82 \times 10^4 \, \text{J}
]
(b) Change in internal energy (( \Delta U )):
This is also directly provided:
[
\Delta U = -4.07 \times 10^4 \, \text{J}
]
(c) Heat (Q):
Using the first law of thermodynamics:
[
\Delta U = Q – W
]
Rearranging the formula to solve for ( Q ):
[
Q = \Delta U + W
]
Substituting the known values:
[
Q = (-4.07 \times 10^4 \, \text{J}) + (1.82 \times 10^4 \, \text{J})
]
[
Q = -2.25 \times 10^4 \, \text{J}
]
Final answers:
- (a) Work, ( W = 1.82 \times 10^4 \, \text{J} ).
- (b) Change in internal energy, ( \Delta U = -4.07 \times 10^4 \, \text{J} ).
- (c) Heat, ( Q = -2.25 \times 10^4 \, \text{J} ).
Explanation:
In this problem, the student is doing work while moving out of the dormitory. According to the first law of thermodynamics, the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Since the student’s internal energy decreases, and we know how much work he does, we can calculate the heat transfer.
Here, the work done is positive, meaning energy is used by the student. Since the internal energy decreases (negative ( \Delta U )), heat must have been lost from the student’s system (negative ( Q )). The negative value for ( Q ) implies that heat flows out of the system during this process. Thus, the student expends energy and loses some heat in the process of doing the work.