Solved thermodynamics exercise

Same change in internal energy, three modes of transfer

Exercise 7 · Lesson 4 — The First Law of Thermodynamics

first law
internal energy
heat
mechanical work
electrical work
calorimetry
state function

Draft version — this exercise is still under review.

Statement

A liquid, its calorimeter, and a small immersed resistor form a closed system that is macroscopically at rest. The container is rigid, and the total heat capacity of the system is assumed constant:

C=2{,}00\ \mathrm{kJ\,K^{-1}}.

The system is to be taken from equilibrium state $A$ , at $T_A=20{,}0\,^\circ\mathrm{C}$ , to equilibrium state $B$ , at $T_B=25{,}0\,^\circ\mathrm{C}$ . It is given that

U(B)-U(A)=C(T_B-T_A).

The changes in macroscopic kinetic and potential energy are zero. All losses to the surroundings are neglected.

The following three protocols are carried out independently, starting from the same initial state $A$ :

Protocol 1. The system is placed in thermal contact with a hot reservoir, with no work supplied to it.
Protocol 2. The walls are thermally insulated. A paddle stirs the liquid as a mass $m$ falls through a height $h=5{,}00$ m. The paddle and liquid end at rest, and all the mechanical energy lost by the mass is transferred to the system. Take $g=9{,}81\ \mathrm{m\,s^{-2}}$ .
Protocol 3. The system receives heat $Q_3=4{,}00$ kJ, and the remaining energy is supplied electrically to the resistor. The electrical power received is constant and equal to $\mathcal P=300$ W.

Calculate the change in internal energy $\Delta U=U(B)-U(A)$ common to all three protocols.
For each of the first two protocols, determine the heat $Q$ and work $W$ received by the system. In the second protocol, calculate the required mass $m$ .
For the third protocol, calculate the electrical work received and the duration for which the resistor is powered.
The three transformations have the same initial and final states. Explain why their values of $Q$ and $W$ may nevertheless differ. Does the system contain “heat” or “work” in the final state $B$ ?

Hint

Apply

\Delta U=Q+W

using the banker's convention. For the paddle, the work received equals the decrease

mgh

in the potential energy of the mass. Electrical energy crossing the boundary through the wires is counted as electrical work.

Detailed solution

Solution

Question 1. The temperature difference is

T_B-T_A=5{,}0

K. Therefore,

\Delta U=C(T_B-T_A) =2{,}00\times5{,}0 =10{,}0\ \mathrm{kJ}.

This value depends only on the two equilibrium states $A$ and $B$ .

Question 2. In the first protocol, no work is received: $W_1=0$ . The first law therefore gives

Q_1=\Delta U=10{,}0\ \mathrm{kJ}.

The heat is positive because energy enters the system.

In the second protocol, the walls are thermally insulated, so $Q_2=0$ . The entire change in internal energy comes from the mechanical work of the paddle:

W_2=\Delta U=10{,}0\ \mathrm{kJ}.

Since $W_2=mgh$ , it follows that

m=\frac{W_2}{gh} =\frac{10{,}0\times10^3}{9{,}81\times5{,}00} \simeq 204\ \mathrm{kg}.

The large order of magnitude is a reminder that even a modest temperature rise already corresponds to a substantial amount of mechanical energy.

Question 3. The first law requires

W_3=\Delta U-Q_3=10{,}0-4{,}00=6{,}00\ \mathrm{kJ}.

This work is positive: the electrical supply provides energy to the system. With $W_3=\mathcal P\,\Delta t$ ,

\Delta t=\frac{W_3}{\mathcal P} =\frac{6{,}00\times10^3}{300} =20{,}0\ \mathrm{s}.

Question 4. The three paths give, respectively,

(Q_1,W_1)=(10{,}0,0)\ \mathrm{kJ}, \qquad (Q_2,W_2)=(0,10{,}0)\ \mathrm{kJ},

and

(Q_3,W_3)=(4{,}00,6{,}00)\ \mathrm{kJ}.

In every case, the sum $Q+W$ is $10{,}0$ kJ and produces the same change $\Delta U$ . Internal energy is a state function, whereas heat and work describe transfers during a transformation. In final state $B$ , the system has internal energy, but it contains neither “heat” nor “work.”