Question

13.During a thermodynamic cycle heat
transfer during a various process is 120
KJ, -20 KJ, -48 KJ and 15 KJ. Find the net
work during the cycle *​

Answer 1

Explanation:

9.5. It is indeed sad that she has lost her father. (Change into Exclamatory)

Answer 2

Given:

The process is cyclic.

Q1 = 120 kJ

Q2 = -20 kJ

Q3 = -48 kJ

Q4 = 15 kJ

To Find:

Net work done during the cycle, W.

Solution:

First law of thermodynamics states that, "the algebraic sum of work and heat tranfers in a cyclic process will be equal".

Net work W = Net heat transfer Q

Net work W = Q1 + Q2 + Q3 + Q4

W = 120 - 20 - 48 + 15

W = 135 - 68

W = 67 kJ.

The net work during the cycle is 67 kJ.