Question

A mass m of a liquid at temperature T₁ is mixed with an equal mass of same liquid temperature T₂. The system is thermally insulated. Show that the entropy change of

Universe is 2mc, log(( T₁+T₂) /2√(T₁T₂) )

Answer 1

Answer:

Given :

A cuboid is 6 cm long, 6 cm wide and 4 cm high .

Edge of cube is 6 cm .

To Find :

Greater volume .

Solution :

Firstly we'll find the Volume of Cuboid :

\longmapsto\tt{Length\:(l)=6\:cm}⟼Length(l)=6cm

\longmapsto\tt{Breadth\:(b)=6\:cm}⟼Breadth(b)=6cm

\longmapsto\tt{Height\:(h)=4\:cm}⟼Height(h)=4cm

Using Formula :

\longmapsto\tt\boxed{Volume\:of\:Cube=l\times{b}\times{h}}⟼

VolumeofCube=l×b×h

Putting Values :

\longmapsto\tt{6\times{6}\times{4}}⟼6×6×4

\longmapsto\tt{36\times{4}}⟼36×4

\longmapsto\tt\bf{144{cm}^{3}}⟼144cm

3

Now ,

For Volume of Cube :

Using Formula :

\longmapsto\tt\boxed{Volume\:of\:Cube={a}^{3}}⟼

VolumeofCube=a

3

Putting Values :

\longmapsto\tt{{6}^{3}}⟼6

3

\longmapsto\tt\bf{216{cm}^{3}}⟼216cm

3

So , The Volume of Cube is greater than the Volume of Cuboid .

Answer 2

Explanation:

Thus change in entropy is given by:

∆S = S2 – S1)

= mc (T1 + T2)/2∫ T1 (dT / T) – mc T2∫ T1 + T2)/2 (dT / T)

= mc ln (T1 + T2)/2 T1) –

mc ln (2 T2) / (T1 + T2)

= mc ln (T1 + T2)/2 T1) +

mc ln (T1 + T2) /2 T2)

= mc ln (T1 + T2) 2 / 4 T1 T2

= mc ln [(T1 + T2) / 2 (T1 T2) 1/2] 2

= 2 mc ln [(T1 + T2) / 2 (T1 T2) 1/2]

= 2 mc ln [(T1 + T2) / 2 ]/[ (T1 T2)1/2]

Hence, Resultant change of entropy of universe is:

2 mc ln [(T1 + T2) / 2]/[ (T1 T2)1/2]

The arithmetic mean (T1 + T2) / 2 is greater than the geometric mean (T1 T2) 1/2

Therefore, ln [(T1 + T2) / 2]/[ (T1 T2)1/2 ]

is always positive. Hence the entropy of the universe increases