Four hundred seventy joules of heat are removed from a heat reservoir at a temperature of 470 K. What is the entropy change of t

Question

3 years ago

11

he reservoir (in J/K)?

2 answers:

11111nata11111 [884] · Answer 1 · 2022-07-03T07:03:17+03:00

Answer:

ΔS= -1 J/K

Explanation:

Given data

Heat Q= -470J

Temperature T=470 K

To find

Entropy change ΔS

Solution

We know that the entropy change of system is ΔS is given by

ΔS=Q/T

We have take heat value Q as negative because the heat is removed from heat reservoir

So

ΔS=(-470J/470K)

ΔS= -1 J/K

irina [24] · Answer 2 · 2022-07-03T08:16:40+03:00

<h2>Answer:</h2>

-1 J/K

<h2>Explanation:</h2>

The entropy (ΔS) of a system is the ratio of the quantity of heat (Q) exchanged (removed or added) in the system to the temperature (T) involved. i.e

ΔS = $\frac{Q}{T}$ ---------------------------(i)

Since, heat was removed from the system the quantity of heat is negative. i.e

Q = -ve

<em>From the question;</em>

Q = -470J

T = 470K

<em>Substitute these values into equation (i) as follows;</em>

ΔS = $\frac{-470}{470}$ = - 1J/K

Therefore, the change in entropy of the system (reservoir in this case) is - 1 J/K