A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
A) What was the heat input to this engine?
B) What was its efficiency?

Respuesta :

Answer:

a) [tex]H_{in}=39.34 kJ[/tex]

b) Efficiency=76.77%

Explanation:

a)

In order to solve this problem, we can use the following formula:

[tex]H_{in}=H_{out}+W[/tex]

the problem provides us with all the necessary information so we can directly use the formula:

[tex]H_{in}=9.14kJ+30.2kJ[/tex]

[tex]H_{in}=39.34 kJ[/tex]

b) In order to find the efficiency, we can use the following formula:

[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]

so we get:

[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]

Efficiency=76.77%