Question

A heat engine operates at 100 C and the room temperature is 23 C. By heating the engine gas, 100 joules of heat is added to a system and at the same time engine does 60 joules of work.

a.Find the maximum efficiency of the engine (5 points)

b.Find the change in internal energy of the gas of the engine :

Answer 1

Answer 2

Given:

• Hot reservoir temperature (T_hot): 100°C
  (Convert to Kelvin: 100 + 273 = 373 K)

• Cold reservoir temperature (T_cold): 23°C
  (Convert to Kelvin: 23 + 273 = 296 K)

• Heat added to the system (Q_in): 100 J

• Work done by the engine (W): 60 J

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Part (a): Maximum Efficiency of the Engine

The maximum possible efficiency (Carnot efficiency) of a heat engine is given by:

$${\eta }_{\text{max}}=1-\frac{T_{\text{cold}}}{T_{\text{hot}}}$$

Plugging in the values:

$${\eta }_{\text{max}}=1-\frac{296\text{ K}}{373\text{ K}}=1-0.7936=0.2064$$

Answer:
The maximum efficiency of the engine is 20.64%.

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Part (b): Change in Internal Energy of the Gas

The First Law of Thermodynamics states:

$$\mathrm{\Delta }U=Q_{\text{in}}-W$$

Where:

• $\mathrm{\Delta }U$ = Change in internal energy

• $Q_{\text{in}}$ = Heat added to the system (100 J)

• $W$ = Work done by the system (60 J)

Plugging in the values:

$$\mathrm{\Delta }U=100\text{ J}-60\text{ J}=40\text{ J}$$

Answer:
The change in internal energy of the gas is 40 J