The Carnot engine's performance can be calculated based on thermodynamic efficiency and energy transfer between the hot and cold reservoirs. Let's solve each part step by step.

Given:

• Temperature of the hot reservoir ($T_h$) = 900 K
• Temperature of the cold reservoir ($T_c$) = 600 K
• Heat absorbed ($Q_h$) = 16000 J

(a) Heat Energy Discarded to Cold Reservoir ($Q_c$)

The efficiency of a Carnot engine is given by:

$\eta = 1 - \frac{T_c}{T_h}$

Rearranging for $Q_c$ using the relation $\eta = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}$:

$Q_c = Q_h \times \frac{T_c}{T_h}$

Substitute the values:

$Q_c = 16000 \times \frac{600}{900} = 16000 \times \frac{2}{3} = 10666.67 \, \text{J}$

So, the heat energy discarded to the cold reservoir is approximately:

$Q_c \approx 10667 \, \text{J}$

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(b) Mechanical Work Performed by the Engine ($W$)

The work $W$ done by the engine is the difference between the heat absorbed and the heat discarded:

$W = Q_h - Q_c$

Using the values we found:

$W = 16000 - 10667 = 5333 \, \text{J}$

So, the mechanical work performed by the engine is approximately:

$W \approx 5333 \, \text{J}$

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(c) Efficiency of the Carnot Engine ($\eta$)

We already have the efficiency formula for a Carnot engine:

$\eta = 1 - \frac{T_c}{T_h}$

Substituting the values:

$\eta = 1 - \frac{600}{900} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.333$

Expressing this as a percentage:

$\eta \approx 33.3\%$

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Summary of Results:

• (a) Heat energy discarded to cold reservoir, $Q_c$: 10667 J
• (b) Mechanical work performed, $W$: 5333 J
• (c) Efficiency of the Carnot engine, $\eta$: 33.3%

Would you like a detailed explanation for any part?

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Related Questions:

1. How does the efficiency of a Carnot engine change if the temperature of the cold reservoir increases?
2. If the temperature difference between the hot and cold reservoirs is doubled, how does it affect the work output?
3. What would be the efficiency of this engine if it operated between 500 K and 800 K instead?
4. How is the Carnot cycle different from a real-life thermodynamic cycle?
5. What factors limit the practical efficiency of a Carnot engine?

Tip:

To improve efficiency in any heat engine, increasing the temperature difference between the hot and cold reservoirs generally helps.