Question

A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00 MJ into the engine. How much heat transfer occurs to the environment? (b) What is the efficiency of the engine?

Answer 1

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:$W=Q=Q_{in}-Q_{out}$ solving to Q_out we get:$Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ)$ this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: $n=1-\frac{T_{Low} }{T_{High}}$ where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:$n=1-\frac{423.15}{723.15} =0.415=41.5%$