Question

Calculate the change in entropy when 20 kg of water is heated from 80°C to 100°C.
(specific heat of water is 4180 J mole-1 K-1)​

Answer 1

Explanation:

ΔS=2.303n×C

p

×log

T

1

T

2

[where,minkgandC

p

inJ/kg]

Entropy change for heating water from 27

o

C to 100

o

C.

ΔS

1

=2.303×

18

1000

×

1000

4180×18

×log

300

373

=910.55J

Entropy change for heating 1 kg H

2

O

(l)

to 1 kg steam at 100

o

C.

ΔS

2

=∫

373

473

T

nC

p

.dT

=m∫

373

473

T

(1670+0.49T)dt

(n is replaced by m since value of C

p

in 3kg

−1

K

−1

m∫

373

473

T

1670dT

+m∫

373

473

0.49dT

m×1670×2.303×[logT]

473

373

+m×0.49[T]

473

373

=1×1670×2.303×log

373

473

+1×0.49×100

=396.73+49=445.73J

Entropy change for heating 1 kg steam at 100

o

C to 1 kg steam at 200

o

C.

ΔS

3

=

T

q

=

373

23×10

5

=6166.2 J K

−1

∴ Total entropy change=ΔS

1

+ΔS

2

+ΔS

3

=910.55+6166.22+445.73

=7522.50J

Answer 2

Answer:

hope it's helpful to you