A certain heat engine does 8.0 kJ of work and dissipates 8.00 kJ of waste heat in a cyclical process.
(a) What was the heat input to this engine?
(b) What was its efficiency?
Hmmm. Wouldn't it be 16kJ input?
wouldn't the efficiency be 50 percent?
To find the answers to these questions, we need to use the first law of thermodynamics, which states that the net heat input into a system is equal to the work done by the system plus the waste heat dissipated.
(a) To find the heat input, we can use the equation:
Heat input = Work done + Waste heat dissipated
Given that the work done is 8.0 kJ and the waste heat dissipated is 8.00 kJ, we can substitute these values into the equation:
Heat input = 8.0 kJ + 8.00 kJ
Therefore, the heat input to the engine is 16.0 kJ.
(b) The efficiency of a heat engine is given by the ratio of the net work done by the engine to the heat input. Mathematically, it can be expressed as:
Efficiency = Work done / Heat input
Substituting the values of work done (8.0 kJ) and heat input (16.0 kJ) into this equation, we can calculate the efficiency:
Efficiency = 8.0 kJ / 16.0 kJ
Simplifying this expression gives us an efficiency of 0.5, or 50%.
Therefore, the (a) heat input to the engine is 16.0 kJ, and (b) its efficiency is 50%.