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Semenov [28]
3 years ago
13

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

Physics
2 answers:
anastassius [24]3 years ago
5 0
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}
cricket20 [7]3 years ago
5 0

ΔS of the object : <u>0.085324 J mol/K</u>

<h3>Further explanation </h3>

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

<h3>∆S ° (reaction) = ∑S ° (product) - ∑ S ° (reagent) </h3>

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}}}

<h3>Learn more  </h3>

Delta H solution  

brainly.com/question/10600048  

an exothermic reaction  

brainly.com/question/1831525  

as endothermic or exothermic  

brainly.com/question/11419458  

an exothermic dissolving process  

brainly.com/question/10541336

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And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

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G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

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The statement that can be used to answer this  question is:

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The potential energy is converted to thermal energy when the object is released the velocity becomes higher because of the acceleration due to gravity.
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A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?
ra1l [238]

Answer:

a=-.78m/s^2

Explanation:

Δv=at

  • Δv is the difference in velocity before and after a given time.
  • a is the acceleration of the object during this time.
  • t is time

(v_f-v_i)=at is another way to write this equation.

  • The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δv=(v_f-v_i)

v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}

  • I rearranged this equation to solve for a, but this is a step that you don't need to take, it's just good to get in the habit of doing this.
  • Plug in the given values. Note that our final velocity is 0, because the car travels until at <em>rest</em>.

a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}

  • Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}

  • if you converted correctly, your answer for v_f will be ≅ 7.6m/s.
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a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2

  • Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
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