Understanding Carnot Engine Efficiency

The efficiency (η) of a Carnot engine operating between two reservoirs at absolute temperatures T₁ (hotter) and T₂ (colder) is given by:

$\eta =1-\frac{T_2}{T_1}$

Step 1: Write the given conditions as equations

Initial efficiency: $\eta =\frac{1}{6}$

So, $\frac{1}{6}=1-\frac{T_2}{T_1}$

Rearranged: $\frac{T_2}{T_1}=1-\frac{1}{6}=\frac{5}{6}$

Thus, $T_2=\frac{5}{6}{T}_1$ (Equation 1)

After T₂ is lowered by 62 K, new T₂' = T₂ - 62

New efficiency: $\eta =\frac{1}{3}$

So, $\frac{1}{3}=1-\frac{T_2-62}{T_1}$

Rearranged: $\frac{T_2-62}{T_1}=1-\frac{1}{3}=\frac{2}{3}$

Thus, $T_2-62=\frac{2}{3}{T}_1$ (Equation 2)

Step 2: Substitute Equation 1 into Equation 2

From Equation 1: $T_2=\frac{5}{6}{T}_1$

Substitute into Equation 2: $\frac{5}{6}{T}_1-62=\frac{2}{3}{T}_1$

Step 3: Solve for T₁

Bring like terms together: $\frac{5}{6}{T}_1-\frac{2}{3}{T}_1=62$

Convert to common denominator (6): $\frac{5}{6}{T}_1-\frac{4}{6}{T}_1=62$

Simplify: $\frac{1}{6}{T}_1=62$

Thus, $T_1=62\times 6=372\text{K}$

Step 4: Solve for T₂ using Equation 1

$T_2=\frac{5}{6}\times 372=310\text{K}$

Final Answer

T₁ = 372 K and T₂ = 310 K

Therefore, the correct option is: 372 K and 310 K

Related Topics and Formulae

Carnot Theorem

No heat engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same reservoirs.

Key Formulae

Carnot efficiency: $\eta =1-\frac{T_c}{T_h}$

Where T_h is the hotter reservoir temperature and T_c is the colder reservoir temperature (in Kelvin)

Important Notes

1. Carnot efficiency represents the maximum possible efficiency for a heat engine

2. All temperatures must be in absolute scale (Kelvin)

3. Efficiency increases as the temperature difference between reservoirs increases