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 .
