Answer to Question #193535 in Molecular Physics | Thermodynamics for Tobias Amukwiita

Question

You make tea with 0.250 kg of 85.0°C water and let it cool to room temperature (20.0°C). (a) Calculate

the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the

air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of

the heat lost by the water goes into the air. What is the total entropy change of the system tea + air?

Answer 1

Gives

$T_1=85°C=273+85=358K$

$T_2=20°C=273+29=293K$

m=0.250kg

Part (a)

$∆S=\smallint_{T_1}^{T_2}\frac{Q}{T}$

Q=mc∆T

∆S=$\smallint_{T_1}^{T_2}\frac{ms∆T}{T}$

∆S=ms$ln\frac{T_2}{T_2}$

Put value

∆S=$0.250\times4190\times ln\frac{293}{358}$

∆S=1047.5$\times ln0.8184$

∆S=1047.5$\times(-0.20036)$

∆S=$-210J/K$

Part(b) channge of entropy

Q=mc∆T

Put value

$$Q=0.250\times4190\times(293-358)J$$

$Q= -6.8\times10^{4}J$