In moving out of a dormitory at the end of the sememster, a student does 1.8 104 J of work. In the process, his internal energy decreases by 4.2 104 J. Determine each of the following quantities (including the algebraic sign).
a) W
b) U
c) Q
To determine each of the quantities (work, internal energy, and heat) in this scenario, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat transferred (Q) into the system minus the work (W) done by the system:
ΔU = Q - W
Given that the internal energy decreases by 4.2 * 10^4 J, we can substitute this value into the equation:
-4.2 * 10^4 J = Q - W
Now, let's solve for each quantity:
a) Work (W):
Since work is the energy transferred due to mechanical displacement, we know that the student does 1.8 * 10^4 J of work while moving out of the dormitory. Therefore, the work done is W = -1.8 * 10^4 J. The negative sign indicates that work is done by the system (the student).
b) Internal energy (U):
As stated, the internal energy decreases by 4.2 * 10^4 J. Hence, the change in internal energy (ΔU) is ΔU = -4.2 * 10^4 J. Again, the negative sign signifies a decrease in internal energy.
c) Heat (Q):
To find the heat transferred (Q), we rearrange the equation as:
Q = ΔU + W
Substituting the given values:
Q = (-4.2 * 10^4 J) + (-1.8 * 10^4 J)
Q = -6 * 10^4 J
Therefore, the heat transferred is Q = -6 * 10^4 J. The negative sign indicates that heat is transferred out of the system.