Respuesta :
Answer:
1) T(final) =238.7K
2) Change in internal energy = 0
3) ΔS = +1.24 J/K
Explanation:
a) Heat lost by Cu (copper) block = Heat gained by lead (Pb)
[tex]-q(Cu) = q(Pb)[/tex]
[tex]-[m(Cu)*c(Cu)*(T2-T1,Cu)] = [m(Pb)*c(Pb)*(T2-T1,Pb)][/tex]
where m = mass of the substance
c = heat capacity
T2 and T1 are the final and initial temperatures
Substituting the appropriate values:
[tex]-[0.032*386*(T2-310)] = [0.0887*128*(T2-161)] \\\\= -[12.352T2-3829.12] = 11.354T2-1827.93\\\\T2 =238.7K[/tex]
b) The change in internal energy between the initial and equilibrium state is zero since the system is insulated as a result of which no work is done.
c) The change in entropy is given as:
[tex]\Delta S = \int \frac{dq}{T}= \int \frac{mc*dT}{T}= mc*ln(\frac{T2}{T1})[/tex]
For the 2 block system:
[tex]\Delta S = \Delta S(Cu) + \Delta S(Pb)= m(Cu)c(Cu)*ln(\frac{T2}{T1})+m(Pb)c(Pb)*ln(\frac{T2}{T1})[/tex]
[tex]= 0.032*386*ln\frac{238.7}{310} +0.0887*128*ln\frac{238.7}{161} \\\\=-3.228 + 4.471 = +1.24 J/K[/tex]
