Question

What is the efficiency of a cyclical heat engine in which 75.0 kJ of heat transfer occurs to the environment for every 95.0 kJ of heat transfer into the engine?

a) 20.0%
b) 30.0%
c) 40.0%
d) 50.0%

Answer 1

The efficiency of a cyclical heat engine can be determined by calculating the ratio of work output to heat input. In this case, the efficiency is approximately 21.05%.

Step-by-step explanation:

The efficiency of a cyclical heat engine is determined by the ratio of work output to heat input. In this case, the heat transfer into the engine is 95.0 kJ and the heat transfer to the environment is 75.0 kJ. To find the efficiency, we divide the work output by the heat input and multiply by 100%.

The work output can be found using the equation: work output = heat input - heat output.

Using the given values, the work output is calculated as: work output = 95.0 kJ - 75.0 kJ = 20.0 kJ.

Now, we can calculate the efficiency: efficiency = (work output / heat input) x 100% = (20.0 kJ / 95.0 kJ) x 100% = 21.05%.

Therefore, the efficiency of the cyclical heat engine is approximately 21.05%.