Question

You make tea with 0.50 kg of 85.0°C water and let it cool to room temperature 120.0°C2.

Answer 1

Answer:

Explanation:

a ) Entropy change dS = dQ/T

= mcdT /T

Integrating both sides

S₂ - S₁ = - mclnT₂ /T₁

= - .5 X 4200 ln (85+273) /( 20 + 273 )

.5 X 4200 ln 358/ 293

= - 417.6 J/K

Entropy change will be negative as heat is lost by the system .

b ) Sine there is no change in the temperature of air , This heat will enter air at temperature ( 20+ 273) K = 293 K

Heat entering air

= .5 x 4200 x 65

= 136500 J

Change in entropy

136500 / 293 ( room temperature is constant at 293k

= + 465.87 J/K

Entropy change will be positive  as heat is gained  by the system .

Total change in the entropy of the system (tea + air )

= +465.87 - 417.6

= 48.27 J/K

Entropy change will be negative as heat is lost by the system .