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Four hundred seventy joules of heat are removed from a heat reservoir at a temperature of 470 K. What is the entropy change of the reservoir (in J/K)?

Respuesta :

Khoso123

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

stigawithfun

Answer:

-1 J/K

Explanation:

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 = [tex]\frac{Q}{T}[/tex]       ---------------------------(i)

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

Q = -ve

From the question;

Q = -470J

T = 470K

Substitute these values into equation (i) as follows;

ΔS = [tex]\frac{-470}{470}[/tex] = - 1J/K

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

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