AdvancedPhysicsMiddle Stage (6–8)Heat & Thermodynamics › Heat Engines, the Second Law and EntropyNumericalJEE Advanced Olympiad

Two identical copper blocks, each of mass 2.0 kg and specific heat 385 J/(kg K), are at 400 K and…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two identical copper blocks, each of mass $2.0\ \text{kg}$ and specific heat $385\ \text{J/(kg K)}$ , are at $400\ \text{K}$ and $225\ \text{K}$ . They are used as the hot and the cold reservoir of a small reversible (Carnot) engine. The engine runs in many tiny cycles, taking heat from the hot block and giving heat to the cold block, until the two blocks reach the same temperature. Nothing else exchanges heat with the blocks.

For one tiny cycle, the engine takes in $dQ_1$ at the hot block's temperature $T_1$ and gives out $dQ_2$ at the cold block's temperature $T_2$ . Use the Carnot relation $dQ_1/T_1 = dQ_2/T_2$ to show that $T_1T_2$ stays the same throughout. What is the final common temperature?
How much work does the engine do in all?
How much heat does the hot block give out, and what is the engine's overall efficiency? Compare it with the efficiency of a Carnot engine between $400\ \text{K}$ and $225\ \text{K}$ .
If the blocks were simply put in contact instead, what would the final temperature be?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Use the Carnot relation for each tiny cycle to find how the two temperatures are linked. Then use energy conservation.

Step 1. The link (a). Each block has heat capacity $C = 2.0 \times 385 = 770\ \text{J/K}$ . In one tiny cycle, $dQ_1 = -C\,dT_1$ and $dQ_2 = C\,dT_2$ . So

\displaystyle \frac{-C\,dT_1}{T_1} = \frac{C\,dT_2}{T_2} \quad\Rightarrow\quad \frac{dT_1}{T_1} + \frac{dT_2}{T_2} = 0 \quad\Rightarrow\quad T_1T_2 = \text{constant}

At the end both are at $T_f$ , so $T_f^2 = 400 \times 225$ and $T_f = 300\ \text{K}$ .

Step 2. Work (b). The hot block gives out $C(400 - 300) = 77000\ \text{J}$ , and the cold block takes in $C(300 - 225) = 57750\ \text{J}$ . The difference is the work:

\displaystyle W = C\,(T_1 + T_2 - 2T_f) = 770 \times (625 - 600)\ \text{J} = 19250\ \text{J}

Step 3. Efficiency (c). $\dfrac{W}{Q_h} = \dfrac{19250}{77000} = 25.0\%$ . A Carnot engine between the starting temperatures would reach $1 - \dfrac{225}{400} = 43.75\%$ , but only at the very start: as the blocks come closer in temperature, each tiny cycle gets less efficient.

Step 4. Just touching (d). With no engine, heat flows straight across and no work is done: $T = \dfrac{400 + 225}{2} = 312.5\ \text{K}$ . It is higher than $300\ \text{K}$ by exactly $\dfrac{W}{2C}$ : the energy that the engine turned into work stays in the blocks as heat.

Answer: (a) $300\ \text{K}$ . (b) $19250\ \text{J}$ . (c) $77000\ \text{J}$ ; $25.0\%$ , against $43.75\%$ for a Carnot engine between the starting temperatures. (d) $312.5\ \text{K}$ .

Check. The geometric mean is always below the ordinary mean, so $W = C(T_1 + T_2 - 2\sqrt{T_1T_2}) = C(\sqrt{T_1} - \sqrt{T_2})^2$ is never negative: an engine can never need work to run between a hot and a cold body.

Watch out. You cannot use one Carnot efficiency for the whole run, because both reservoir temperatures keep changing. That is why the work comes from the product rule, not from $\eta Q$ .

← All Problems

← A sample of a monatomic ideal gas is taken from state A, with volume 1.0 L and pressure 3P₀, to…Next →

More onHeat & Thermodynamics › Heat Engines, the Second Law and Entropy

Discussion (0)

Loading…