A reversible cyclic process involves 6 steps. In step -1 and 3 system absorbs 500 J, 800 J of heat from a heat reservoir at temperature 250 K and 200 K respectively. Step 2, 4, 6 are adiabatic such that the temperature of one reservoir changes to that of next. Total work done by the system in whole cycle is 700 J. Find the temperature during step 5 if it exchanges heat from a reservoir at temperature $T_{5}$ .

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 100.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let heat involved in step 5 is $Q_{5}$

$\begin{array}{l} q_{total} = - W_{total} \\ 500 + 800 + Q_{5} = 700 \\ Q_{5} = - 600 J \\ Since Δ S total = 0 \\ \frac{500}{250} + \frac{800}{200} - \frac{600}{T_{5}} = 0 \\ \Rightarrow T_{5} = 100 K \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th