Statement

Two rigid blocks A and B have constant total heat capacities $C_A>0$ and $C_B>0$ . Their initial temperatures satisfy $T_{A,0}>T_{B,0}>0$ . Each block remains at a uniform temperature, which may change as it exchanges heat. No phase change occurs.

A heat engine operates between the blocks, without bringing them into direct contact. It extracts heat from A, transfers part of it to B and delivers work to the surroundings. The combined blocks and engine exchange no heat with the surroundings. The engine returns to the same state after each cycle: it accumulates neither energy nor entropy. We seek the maximum work recoverable from the initial temperature difference.

Consider the limit of a succession of infinitesimal reversible cycles. During each cycle, the instantaneous temperatures $T_A$ and $T_B$ change infinitesimally. Use the received-energy sign convention: energy is positive when received by the system considered. Denote by $\delta Q_A$ and $\delta Q_B$ the heats received by the engine from A and B respectively, and by $\delta W$ the work received by the engine. During engine operation, $\delta Q_A>0$ , $\delta Q_B<0$ and $\delta W<0$ . The heats received by the blocks are therefore $-\delta Q_A$ and $-\delta Q_B$ respectively.

Write the energy balances for the blocks and the engine over an infinitesimal cycle. Use the Carnot efficiency to express $\delta Q_B$ in terms of $\delta Q_A,T_A,T_B$ . Show that A cools and B warms. How do the temperature difference and efficiency change over successive cycles? Under what condition does work production cease?
Establish a relation between $\mathrm dT_A/T_A$ and $\mathrm dT_B/T_B$ , then integrate it from the initial state. Deduce the common final temperature $T_r$ . Why can the blocks not be treated as reservoirs at constant temperatures $T_{A,0}$ and $T_{B,0}$ ?
Calculate the total work received $W_{\mathrm{rev}}$ during this reversible operation, then its absolute value $|W_{\mathrm{rev}}|$ , the maximum work recoverable outside the system. Simplify $T_r$ and $W_{\mathrm{rev}}$ when $C_A=C_B=C$ ; express $|W_{\mathrm{rev}}|$ in a manifestly positive form.
Numerical application: both blocks are iron, each of mass $m=1{,}0\ \mathrm{kg}$ , initially at $60\,{}^\circ\mathrm C$ for A and $20\,{}^\circ\mathrm C$ for B. Assume a constant specific heat capacity for iron, $c=450\ \mathrm{J\,kg^{-1}\,K^{-1}}$ . Calculate the maximum recoverable work $|W_{\mathrm{rev}}|$ .

Hint

Over each infinitesimal cycle, the efficiency is

1-T_B/T_A

, using the instantaneous temperatures, and equals

|\delta W|/\delta Q_A

. For the overall balance, apply

\Delta U=Q+W

to the combined blocks and engine, with

Q=0

and

W<0

.

Detailed solution

Solution

Question 1. The blocks are rigid and receive no work. Their energy changes are therefore the heats received:

C_A\,\mathrm dT_A=-\delta Q_A,\qquad C_B\,\mathrm dT_B=-\delta Q_B.

The engine returns to its initial state. Its energy balance gives $0=\delta Q_A+\delta Q_B+\delta W$ , or

\delta W=-(\delta Q_A+\delta Q_B)<0.

Efficiency is the ratio of recovered work to heat received from A. Since $\delta W<0$ , we have $|\delta W|=-\delta W$ , so

\eta=\frac{|\delta W|}{\delta Q_A} =1+\frac{\delta Q_B}{\delta Q_A}=1-\frac{T_B}{T_A} \quad\Longrightarrow\quad \delta Q_B=-\frac{T_B}{T_A}\delta Q_A.

Block A receives $-\delta Q_A<0$ and block B receives $-\delta Q_B>0$ . With positive heat capacities, $\mathrm dT_A<0$ and $\mathrm dT_B>0$ . The temperature difference therefore decreases:

\mathrm d(T_A-T_B) =-\delta Q_A\left(\frac1{C_A}+\frac{T_B}{C_BT_A}\right)<0.

The ratio $T_B/T_A$ increases because its numerator increases while its denominator decreases. The efficiency $1-T_B/T_A$ gradually falls. When $T_A=T_B$ , it vanishes: the temperature difference is exhausted, and no cyclic engine operating between these blocks can deliver further work.

Question 2. The block balances give

\delta Q_A=-C_A\,\mathrm dT_A

and

\delta Q_B=-C_B\,\mathrm dT_B

. Substitute into the previous relation:

-C_B\,\mathrm dT_B =-\frac{T_B}{T_A}(-C_A\,\mathrm dT_A) =\frac{T_B}{T_A}C_A\,\mathrm dT_A.

Changing the sign and dividing by $T_B$ gives

C_B\,\mathrm dT_B=-\frac{T_B}{T_A}C_A\,\mathrm dT_A, \qquad C_A\frac{\mathrm dT_A}{T_A} +C_B\frac{\mathrm dT_B}{T_B}=0.

Integrate from the initial temperatures to those reached after any number of cycles:

C_A\int_{T_{A,0}}^{T_A}\frac{\mathrm dT}{T} +C_B\int_{T_{B,0}}^{T_B}\frac{\mathrm dT}{T}=0.

An antiderivative of $1/T$ is $\ln T$ ; evaluation at the limits gives

C_A\ln\frac{T_A}{T_{A,0}}+C_B\ln\frac{T_B}{T_{B,0}}=0.

This relation expresses conservation of total entropy during reversible operation. When the engine stops, $T_A=T_B=T_r$ . Writing $\ln(T_r/T_{B,0})=\ln(T_r/T_{A,0})+\ln(T_{A,0}/T_{B,0})$ gives

(C_A+C_B)\ln\frac{T_r}{T_{A,0}} =C_B\ln\frac{T_{B,0}}{T_{A,0}}.

Dividing by $C_A+C_B$ and exponentiating gives

\boxed{T_r=T_{A,0} \left(\frac{T_{B,0}}{T_{A,0}}\right)^{C_B/(C_A+C_B)}.}

The blocks have finite heat capacities: any heat exchange changes their temperatures according to $\mathrm dT=\delta Q_{\mathrm{block}}/C$ , where $\delta Q_{\mathrm{block}}$ is the heat received by the block considered. The initial efficiency therefore does not apply throughout the experiment. Constant temperatures can be assumed only within each infinitesimal cycle, using their updated values each time.

Question 3. The combined blocks and engine are adiabatic but deliver work: they are not isolated. Since the engine stores no energy, the first law gives

C_A(T_r-T_{A,0})+C_B(T_r-T_{B,0})=W_{\mathrm{rev}}.

The work received by the combined system is that received by the engine. Collecting terms gives

W_{\mathrm{rev}}=(C_A+C_B)T_r-C_AT_{A,0}-C_BT_{B,0}<0.

The work recovered outside the system is its absolute value:

\boxed{|W_{\mathrm{rev}}|=-W_{\mathrm{rev}} =C_AT_{A,0}+C_BT_{B,0}-(C_A+C_B)T_r.}

Each product $CT$ has the unit of energy, the joule. To see why reversibility gives the maximum, consider an engine that produces entropy and leaves the blocks at a common temperature $T_*$ . The blocks' entropy balance requires

C_A\ln\frac{T_*}{T_{A,0}}+C_B\ln\frac{T_*}{T_{B,0}}\geq0.

Subtracting the reversible balance, which is zero, leaves $(C_A+C_B)\ln(T_*/T_r)\geq0$ , hence $T_*\geq T_r$ . The work received during this engine operation is $W_*=(C_A+C_B)T_*-C_AT_{A,0}-C_BT_{B,0}$ , so

W_*-W_{\mathrm{rev}}=(C_A+C_B)(T_*-T_r)\geq0.

Thus $W_{\mathrm{rev}}\leq W_*\leq0$ : the work received is most negative in the reversible case. Its absolute value is therefore maximal, $|W_*|\leq|W_{\mathrm{rev}}|$ .

For

C_A=C_B=C

, the exponent is

1/2

, hence

T_r=T_{A,0}\sqrt{\frac{T_{B,0}}{T_{A,0}}} =\sqrt{T_{A,0}T_{B,0}}, W_{\mathrm{rev}}=-C\left(\sqrt{T_{A,0}}-\sqrt{T_{B,0}}\right)^2<0, |W_{\mathrm{rev}}|=C\left(T_{A,0}+T_{B,0} -2\sqrt{T_{A,0}T_{B,0}}\right) =C\left(\sqrt{T_{A,0}}-\sqrt{T_{B,0}}\right)^2>0.

We recognize the identity $a^2+b^2-2ab=(a-b)^2$ . The work vanishes if the initial temperatures are equal.

Question 4. The given heat capacity is per unit mass. The total heat capacity of each block is therefore

C=mc=1{,}0\times450=450\ \mathrm{J\,K^{-1}}.

Temperatures inside the square roots must be expressed in kelvins:

T_{A,0}=60+273{,}15=333{,}15\ \mathrm K,\qquad T_{B,0}=20+273{,}15=293{,}15\ \mathrm K.

Substituting into the result of question 3 gives

|W_{\mathrm{rev}}| =450\left(\sqrt{333{,}15}-\sqrt{293{,}15}\right)^2 \simeq575\ \mathrm J=0{,}575\ \mathrm{kJ}.

The maximum recoverable work is therefore about $0{,}58\ \mathrm{kJ}$ . With the received-energy sign convention, the work received by the engine is $W_{\mathrm{rev}}\simeq-575\ \mathrm J$ : the negative sign means that it transfers this energy to the surroundings.