Question

When an aluminum bar is connected between a hot reservoir at 720 K and a cold reservoir at 358 K, 3.00 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. (a) In this irreversible process, calculate the change in entropy of the hot reservoir._______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?

Answer 1

Answer:

a.  -4.166 J/K

b. 8.37 J/K

c. 4.21 J/K

d. entropy always increases.

Explanation:

Given :

Temperature at hot reservoir , $T_h$ = 720 K

Temperature at cold reservoir , $T_c$ = 358 K

Transfer of heat, dQ = 3.00 kJ = 3000 J

(a). In the hot reservoir, the change of entropy is given by:

$dS_h= -\frac{dQ}{t_h}$              (the negative sign shows the loss of heat)

$dS_h= -\frac{3000}{720}$

      =  -4.166 J/K

(b)  In the cold reservoir, the change of entropy is given by:

$dS_c= \frac{dQ}{t_c}$              

$dS_c= \frac{3000}{358}$

      =  8.37 J/K

(c). The entropy change in the universe is given by:

$dS=dS_h+dS_c$

    = -4.16+8.37

   = 4.21 J/K

(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.