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The efficiency of a carnot engine is (1)...

The efficiency of a carnot engine is `(1)/(6)`. If the temperature of the sink is reduced by 62 K, the efficiency becomes `(1)/(3)`. The temperature of the source and the sink in the first case are respectively.

A

372 K, 290 K

B

372 K, 310 K

C

744 K, 310 K

D

744 K, 290 K

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To solve the problem, we need to find the temperatures of the source (T2) and sink (T1) of a Carnot engine given its efficiencies at two different states. ### Step-by-Step Solution: 1. **Understanding Efficiency of Carnot Engine**: The efficiency (η) of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_1}{T_2} \] where \( T_1 \) is the temperature of the sink and \( T_2 \) is the temperature of the source. 2. **Setting Up the First Equation**: Given that the efficiency is \( \frac{1}{6} \): \[ 1 - \frac{T_1}{T_2} = \frac{1}{6} \] Rearranging gives: \[ \frac{T_1}{T_2} = 1 - \frac{1}{6} = \frac{5}{6} \] Thus, we can express \( T_1 \) in terms of \( T_2 \): \[ T_1 = \frac{5}{6} T_2 \quad \text{(Equation 1)} \] 3. **Setting Up the Second Equation**: When the sink temperature is reduced by 62 K, the new efficiency becomes \( \frac{1}{3} \): \[ 1 - \frac{T_1 - 62}{T_2} = \frac{1}{3} \] Rearranging gives: \[ \frac{T_1 - 62}{T_2} = 1 - \frac{1}{3} = \frac{2}{3} \] Thus, we can express \( T_1 - 62 \) in terms of \( T_2 \): \[ T_1 - 62 = \frac{2}{3} T_2 \quad \text{(Equation 2)} \] 4. **Substituting Equation 1 into Equation 2**: Substitute \( T_1 = \frac{5}{6} T_2 \) into Equation 2: \[ \frac{5}{6} T_2 - 62 = \frac{2}{3} T_2 \] 5. **Solving for T2**: To eliminate the fractions, multiply the entire equation by 6: \[ 5 T_2 - 372 = 4 T_2 \] Rearranging gives: \[ 5 T_2 - 4 T_2 = 372 \] Thus: \[ T_2 = 372 \, \text{K} \] 6. **Finding T1**: Now substitute \( T_2 = 372 \, \text{K} \) back into Equation 1: \[ T_1 = \frac{5}{6} \times 372 = 310 \, \text{K} \] ### Final Answer: The temperatures of the source and sink in the first case are: - Source temperature \( T_2 = 372 \, \text{K} \) - Sink temperature \( T_1 = 310 \, \text{K} \)

To solve the problem, we need to find the temperatures of the source (T2) and sink (T1) of a Carnot engine given its efficiencies at two different states. ### Step-by-Step Solution: 1. **Understanding Efficiency of Carnot Engine**: The efficiency (η) of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_1}{T_2} ...
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Knowledge Check

  • A Carnot engine efficiency is equal to 1/7 . If the temperature of the sink is reduced by 65 K , the efficiency becomes 1/4 . The temperature of the source and the sink in the first case are respectively

    A
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    B
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    C
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    D
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    A
    372, 310
    B
    472, 410
    C
    310, 372
    D
    744, 682
  • A cannot engine has efficiency (1)/(6) . If temperature of sink is decreased by 62^(@)C then its efficiency becomes (1)/(3) then the temperature of source and sink:

    A
    `T_(H)=98^(@)C,T_(C)=36^(@)C`
    B
    `T_(H)=99^(@)C,T_(C)=37^(@)C`
    C
    `T_(H)=100^(@)C,T_(C)=38^(@)C`
    D
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