To solve the problem step by step, we will calculate the efficiency of the Carnot engine and the useful work done per cycle by the engine.
### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures given in the problem are:
- T1 (hot reservoir) = 627 °C
- T2 (cold reservoir) = 27 °C
To convert these temperatures to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Calculating:
- \( T1 = 627 + 273 = 900 \, K \)
- \( T2 = 27 + 273 = 300 \, K \)
### Step 2: Calculate the efficiency of the Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T2}{T1} \]
Substituting the values we found:
\[ \eta = 1 - \frac{300}{900} \]
Calculating:
\[ \eta = 1 - \frac{1}{3} = \frac{2}{3} \]
Converting to percentage:
\[ \eta = \frac{2}{3} \times 100 \approx 66.67\% \]
### Step 3: Calculate the heat exhausted (Q2)
We know that the heat absorbed (Q1) is 1000 kcal. We can use the relation between heat and temperature for the Carnot engine:
\[ \frac{Q1}{Q2} = \frac{T1}{T2} \]
From this, we can derive Q2:
\[ Q2 = Q1 \times \frac{T2}{T1} \]
Substituting the known values:
\[ Q2 = 1000 \, \text{kcal} \times \frac{300 \, K}{900 \, K} \]
Calculating:
\[ Q2 = 1000 \, \text{kcal} \times \frac{1}{3} = \frac{1000}{3} \, \text{kcal} \approx 333.33 \, \text{kcal} \]
### Step 4: Calculate the useful work done (W)
The useful work done by the engine in one cycle can be calculated using:
\[ W = Q1 - Q2 \]
Substituting the values:
\[ W = 1000 \, \text{kcal} - \frac{1000}{3} \, \text{kcal} \]
Calculating:
\[ W = 1000 - 333.33 = 666.67 \, \text{kcal} \]
### Step 5: Convert the work done to Joules
To convert kcal to Joules, we use the conversion factor \( 1 \, \text{kcal} = 4184 \, \text{J} \):
\[ W = 666.67 \, \text{kcal} \times 4184 \, \text{J/kcal} \]
Calculating:
\[ W \approx 666.67 \times 4184 \approx 2.8 \times 10^6 \, \text{J} \]
### Final Results
- Efficiency of the Carnot engine: **66.67%**
- Useful work done per cycle: **666.67 kcal** or approximately **2.8 x 10^6 J**
