A total of 6.28 x 10^5 J is taken from a heat source at 150.0 degrees Celsius and rejected to a heat sink at 50.0 degrees Celsius. Find (a) the change of entropy of in the heat source and heat sink separately and (b) the entropy change in the total system.

Explanation:

a)

change in entropy of source = heat taken / temperature of source

= 6.28 x 10⁵ / (273 + 150)

= 628000 / 423

= 1484.63 JK⁻¹  ( decrease in entropy )

heat rejected in sink = 628000 x  (273 + 50) / 423

= 628000 x  323 / 423

= 479536.64 J

change in entropy of sink  = heat rejected  / temperature of sink

= 479536.64 / (273 + 50)

=479536.64 / 323

= 1484.63 JK⁻¹  ( increase in entropy )

b )

net entropy change in the system

= +1484.63 - 1484.63

= zero.

D. liquid with high the state Viscosity

Explanation:

It still a liquid because its able to take shape of an object, But due to its thickness it hard to tell. So Its counted as a Liquid with high Viscosity ( the state of being thick, sticky, and semifluid in consistency, due to internal friction.)