Question

Heat in the amount of 100 KJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600K.Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.

Answer 1

0.0837 kJ/K

Step-by-step explanation:

Given:

Temperature of the cold reservoir T,cold = 600 K

Temperature of the hot reservoir T,hot= 1200 K

Heat transferred , Q=100 kJ

Now the entropy change for the cold reservoir

`\bigtriangleup S,cold=-(Q)/(T,cold)`

`\bigtriangleup S,cold=-(-100)/(600)`

`\bigtriangleup S,cold=0.1667 kJ/K`

Now the entropy change for the cold reservoir

`\bigtriangleup S,hot=-(Q)/(T,hot)`

`\bigtriangleup S,hot=-(100)/(600)`

`\bigtriangleup S,hot=-0.0833 kJ/K`

Therefore, the total entropy change for the two reservoir is

`\bigtriangleup S=\bigtriangleup S,hot +\bigtriangleup S,cold`

thus,

ΔS=0.1667-0.0833

ΔS=0.0833 kJ/K

Since, the change of entropy is positive thus we can say it is possible and

increase of entropy principle is satisfied

Answer 2

0.0833 k J/k

Step-by-step explanation:

Given data in question

total amount of heat transfedded (Q) = 100 KJ

hot reservoir temperature R(h) = 1200 K

cold reservoir temperature R(c) = 600 k

Solution

we will apply here change of entropy (Δs) formula

Δs =
`(Q)/(R(h))+(Q)/(R(c))`

Δs =
`(-100)/(1200)+(100)/(600))`

Δs =
`(1)/(12)`

Δs = 0.0833 K J/k

this change of entropy Δs is positive so we can say it is feasible and

increase of entropy principle is satisfied