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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 227C{{227}^{\circ }}C; the heat is rejected at 27C{{27}^{\circ }}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

Explanation

Solution

We have the temperature of sink as:T1=270C{{T}_{1}}={{270}^{\circ }}C and temperature of source as: T2=27C{{T}_{2}}={{27}^{\circ }}C. Also, heat supplied is: Q1=250 kJs1{{Q}_{1}}=250\text{ }kJ{{s}^{-1}}. So, by using the formula of efficiency of the heat engine, find the heat rejected: Q2{{Q}_{2}}.

Formula used:
η=1T1T2=1Q1Q2\eta =1-\dfrac{{{T}_{1}}}{{{T}_{2}}}=1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}}, where η\eta is the efficiency of heat engine, T1{{T}_{1}} is the initial temperature, T2{{T}_{2}} is the final temperature, Q1{{Q}_{1}} is the heat supplied and Q2{{Q}_{2}} is the heat rejected.

Complete step by step answer:
We have:
T1=227C T2=27C Q1=250 kJs1 \begin{aligned} & {{T}_{1}}={{227}^{\circ }}C \\\ & {{T}_{2}}={{27}^{\circ }}C \\\ & {{Q}_{1}}=250\text{ }kJ{{s}^{-1}} \\\ \end{aligned}
Now, convert all the temperatures from degree Celsius to degree Kelvin.
We have:
T1=227+273 =500K\begin{aligned} & {{T}_{1}}=227+273 \\\ & =500K \end{aligned}
T2=27+273 =300K\begin{aligned} & {{T}_{2}}=27+273 \\\ & =300K \end{aligned}
Now, by using the formula for efficiency of heat engine, we have:
η=1T2T1 =1500300 =300500300 =200300 =23......(1)\begin{aligned} & \eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}} \\\ & =1-\dfrac{500}{300} \\\ & =\dfrac{300-500}{300} \\\ & =-\dfrac{200}{300} \\\ & =-\dfrac{2}{3}......(1) \end{aligned}
Also, we know that, the efficiency of heat engine is equal to:
η=1Q1Q2\eta =1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}}
Now, by substituting the value of efficiency of the heat engine from equation (1), find the value of heat rejected.
We get:
23=1250Q2 250Q2=1+23 250Q2=53 Q2=250×35 Q2=150kJs1 \begin{aligned} & \Rightarrow -\dfrac{2}{3}=1-\dfrac{250}{{{Q}_{2}}} \\\ & \Rightarrow \dfrac{250}{{{Q}_{2}}}=1+\dfrac{2}{3} \\\ & \Rightarrow \dfrac{250}{{{Q}_{2}}}=\dfrac{5}{3} \\\ & \Rightarrow {{Q}_{2}}=\dfrac{250\times 3}{5} \\\ & \Rightarrow {{Q}_{2}}=150kJ{{s}^{-1}} \\\ \end{aligned}
So, the amount of heat rejected by the heat engine is 150kJs1150kJ{{s}^{-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.