Question
Question: A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of \({{227}^{\circ }...
A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of 227∘C; the heat is rejected at 27∘C, the cycle is reversible, then what amount of heat is rejected?
a) 24 kJ/s
b) 223 kJ/s
c) 150 kJ/s
D) None of the above
Solution
We have the temperature of sink as:T1=270∘C and temperature of source as: T2=27∘C. Also, heat supplied is: Q1=250 kJs−1. So, by using the formula of efficiency of the heat engine, find the heat rejected: Q2.
Formula used:
η=1−T2T1=1−Q2Q1, where η is the efficiency of heat engine, T1 is the initial temperature, T2 is the final temperature, Q1 is the heat supplied and Q2 is the heat rejected.
Complete step by step answer:
We have:
T1=227∘CT2=27∘CQ1=250 kJs−1
Now, convert all the temperatures from degree Celsius to degree Kelvin.
We have:
T1=227+273=500K
T2=27+273=300K
Now, by using the formula for efficiency of heat engine, we have:
η=1−T1T2=1−300500=300300−500=−300200=−32......(1)
Also, we know that, the efficiency of heat engine is equal to:
η=1−Q2Q1
Now, by substituting the value of efficiency of the heat engine from equation (1), find the value of heat rejected.
We get:
⇒−32=1−Q2250⇒Q2250=1+32⇒Q2250=35⇒Q2=5250×3⇒Q2=150kJs−1
So, the amount of heat rejected by the heat engine is 150kJs−1 .
So, the correct answer is “Option C”.
Additional Information:
A heat engine is a device that converts heat to work. It takes heat from a reservoir then does some work like moving a piston, lifting weight etc and finally discharges some heat energy into the sink.
Note:
Remember to convert both the given temperatures in Kelvin scale first. The temperature can be converted by adding 273 to each given value of temperature.