Question

A carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

• A. 200°C
• B. 27°C
• C. 15°C
• D. 100°C

Answer 1

The Correct Option is B

The efficiency of a Carnot engine is given by the formula: 

η=1−(T_c/T_h)

 

where η is the efficiency, T_c is the temperature of the cold reservoir (sink), and T_h is the temperature of the hot reservoir (source). All temperatures are in Kelvin. Given that the efficiency η is 50%, this can be represented as 0.5. The temperature of the hot reservoir T_h is 327°C, which converts to Kelvin as follows:

T_h=327+273=600 K

Substituting the values in the efficiency equation:

0.5=1−(T_c/600)

 

Simplifying the equation gives:

0.5=(T_c/600)

 

Solve for T_c:

T_c=0.5×600=300 K

 

Convert the temperature back to Celsius:

T_c=300−273=27°C

 

Thus, the temperature of the sink is 27°C.

Answer 2

The correct option is (B): 27°C

The efficiency of Carnot engine,

%\(\eta\)=(1-\( \frac{T_{sink}}{T_{source}}\times100\))

\(T_{source}=327^{\circ}C=600K\)

\(50=(1- \frac{T_{sink}}{600})\times100\)

\( \frac{1}{2}=1- \frac{T_{sink}}{600}\)

\(T_{sink}=300K\)

So the temperature of the sink is= 327-300=27\(^{\circ}\)C