Question

What is the work output of a cyclical heat engine with a 22.0% efficiency and 6.00 × 10⁹ J of heat transfer into the engine?

a) 1.32 × 10⁹ J
b) 1.62 × 10⁹ J
c) 1.86 × 10⁹ J
d) 2.04 × 10⁹ J

Answer 1

The work output of a cyclical heat engine with 22.0% efficiency and 6.00×10⁹ J of heat input is 1.32×10⁹ J, and the heat transferred to the environment is 4.68×10⁹ J.

Step-by-step explanation:

The question asks about the work output of a cyclical heat engine with a given efficiency and amount of heat transfer into the engine. Given a 22.0% efficiency and 6.00×10⁹ J of heat transfer into the engine, the first step is to calculate the work output using the formula for efficiency: efficiency = (work output) / (heat input). Thus, work output = efficiency × heat input. Plugging in the given values, we get work output = 0.22 × 6.00×10⁹ J = 1.32×10⁹ J.

Part (b) of the question asks about the heat transfer to the environment. The heat transferred to the environment is the difference between the heat input and the work output. Therefore, heat transfer to the environment = heat input - work output = 6.00×10⁹ J - 1.32×10⁹ J, which equals 4.68×10⁹ J.