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

Answer:

0.0833 k J/k

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 = $\frac{Q}{R(h)}+\frac{Q}{R(c)}$

Δs = $\frac{-100}{1200}+\frac{100}{600)}$

Δs = $\frac{1}{12}$

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

Answer 2

Answer:

0.0837 kJ/K

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

$\bigtriangleup S,cold=-\frac{Q}{T,cold}$

$\bigtriangleup S,cold=-\frac{-100}{600}$

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

Now the entropy change for the cold reservoir

$\bigtriangleup S,hot=-\frac{Q}{T,hot}$

$\bigtriangleup S,hot=-\frac{100}{600}$

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

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