Question

2.
A hot body at 1000 K transfers 2000 kJ of heat to a body at 500 K. Determine the net
entropy change?
(a) +4 kJ/kg
(b) –2 kJ/kg
(c) +2 kJ/kg
(d) -4 kJ/kg​

Answer 1

Sign: Heat released is taken as - ve and heat added to a system is +ve.

del S = (-Q/T1) + (Q/T2)

=(2000/1000) + (2000/500)

=-2 +4

=+2 KJ

Answer 2

Step-by-step explanation:

Given A hot body at 1000 K transfers 2000 kJ of heat to a body at 500 K. Determine the net  entropy change?

• Given a hot body at 1000 K transfers 2000 KJ of heat to a body at 500 K. We need to find the net entropy change.
• Heat transfers from A to B .
• We know that Δ S A = - 2000 KJ / 1000 K
•                                    = - 2
• Now Δ S B = 2000 KJ / 500 K = 4
• So Δ S net = - 2 + 4
•                   = + 2

So net entropy change will be + 2 kJ / Kg

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