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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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amid [387]

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

L_0 = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67

Length contraction is given by

L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m

The length is 85.62 m

Time taken

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years

The time taken from the perspective of the ship is 101.04 years

8 0
3 years ago
A boy in a wheelchair (total mass 54.5 kg) has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom
babymother [125]

Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

initial speed (Vi) = 1.4 m/s

final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

height of slope (h) = 2.1 m

length of slope (d) = 12.4 m

acceleration due to gravity (g) = 9.8 m/s^2

work done (wd) = ?

  • we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy

wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd =  (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)            

wd = 630.75 j

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3 years ago
On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.
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This is a vector subraction problem.

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Answer:

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kogti [31]

Answer:

A) The resultant force is 43.4 [N]

B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis

Explanation:

We need to make a sketch of the different forces acting on the heavy crate.

In the attached image we can see the forces and the sum of the vector with their respective angles.

Forces in the X-axis

Fdionx=18.5N\\\\Fshix=16.5*cos(30)=14.29N\\Fjoanx=19.5*cos(60)=9.75N\\\\Forcex= 18.5 + 14.29 + 9.75 = 42.54 N

Forces in the y-axis

FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]

Using the Pythagorean theorem

Tforce=\sqrt{(42.54)^{2} +(8.63)^{2} } \\\\Tforce= 43.4N

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.

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