Answer to Question #193621 in Molecular Physics | Thermodynamics for Johannes Steven

Question

(a) You make tea with 0.250 kg of 85.0 degrees Celsius water and let it cool to room temperature (20.0 degrees Celsius) (i) Calculate the entropy change of the water while it cools. (ii) 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?

(b) An ice making machine operates in a Carnot cycle. It takes heat from water at 0.0 degrees Celsius and rejects heat to room at 24.0 degrees Celsius. Suppose that 85.0 kg of water at 0.0 _C are converted to 0.0 degrees Celsius. (i) How much heat is discharged into the room? (ii) How much energy must be supplied to the device?

Answer 1

a)

i)

given

m= 0.25 kg , T₁=85^oC =358 K

T₂(cooled temp)=20^o C = 293 K

$\Delta S (entropy \ change )\ = \int_{T_1}^{T_2}\frac{Q}{T} \ \ \ \ \ \ (Q=mcdt)\\ \Delta S = mc\int_{T_1}^{T_2}\frac{dt}{T} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\int\frac{1}{T}=lnT)\\ \Delta S =mcln(\frac{T_2}{T_1})=0.250kg\times J/kg.K \ ln(\frac{293}{358})=\boxed{-210J/K}$

as shown the entropy decreases due to cooling

ii)

$Q=mc\Delta T\\ Q=mc(T_2-T_1) = 0.25kg\times4190 J/kg .K(293K-358K) \\ Q=-6.8\times10^4J$

heat is negative , we are given in this part , the process is isothermal ,

$\Delta S_{air} = \frac{Q}{T}={6.8\times10^4J\over293K}= \boxed{232.1 \ J/K}$

the change in entropy $\Delta S_{net}=\Delta S_{tea}+\Delta S_{air}=-210 \ J/K + 232.1 \ J/K\\ \Delta S_{net}=\boxed{22.1 \ J/K}$

b)

i)

Q_C = heat from ice , Q_H heat discards to the room

L_f= heat fusion =$334\times 10^3 J/kg$

$Q_c=mL_f=85kg\times334\times 10^3 J/kg=1.84\times10^4kJ$

now

$\frac{Q_C}{Q_H}=-\frac{T_C}{T_H}\\ {Q_C}=-{Q_H}\frac{T_C}{T_H}=-(2.84\times10^4kJ)(\frac{297K}{273K})\\ Q_c=\boxed{-3.09\times10^4kJ}$

the negative sign is due to discarding the heat to the hot reservoir

ii)

$|Q_H|=Q_c+|W| \\ |W|=Q_C-|Q_H|=2.84\times10^4-3.09\times10^3kJ\\ W=-2.50\times10^3kJ$

The work is negative which means takes the engine energy equals $\boxed{2.50\times10^3kJ}$