Question

An object at 20∘c absorbs 25.0 j of heat. what is the change in entropy δs of the object?

Answer 1

From the definition of entropy, the entropy change of an object is
$\delta S = \frac{Q}{T}$
where
Q is the heat absorbed
T is the absolute temperature

in our problem, we have Q=25.0 J, while the absolute temperature is (converting in Kelvin)
$T=20 ^{\circ}C + 273 = 293 K$

and so the entropy change is
$\delta S= \frac{25.0 J}{293 K}=0.085 JK^{-1}$

Answer 2

ΔS of the object : 0.085324 J mol/K

Further explanation

Entropy (with the symbol S) in thermodynamics indicated the degree of system disorder

Entropy like other thermodynamic terms can only be calculated from the initial and final changes

Entropy will also change in the process of changing energy forms

The value of ΔS ° can be calculated from standard entropy data

∆S ° (reaction) = ∑S ° (product) - ∑ S ° (reagent)

Entropy can also be calculated from the change in heat (ΔQ) to temperature (T), under Isothermal process (constant temperature) conditions

$\rm \boxed{\Delta S=\dfrac{\Delta Q}{T}}$

From this equation it can be seen that entropy will rise if the temperature gets lower (hot to cold)

An object at 20°C absorbs 25.0 j of heat

T = 20 + 273 = 293 K

ΔQ = 25 J

then:

Entropy: ΔS =

$\rm \Delta S=\dfrac{25\:J}{293\:K}=\boxed{\bold{0.08532\:J\frac{mol}{K}}}$

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