Question

A 32.0 g block of copper whose temperature is 310 K is placed in an insulating box with a 88.7 g block of lead whose temperature is 161 K. (a) What is the equilibrium temperature of the two-block system? (b) What is the change in the internal energy of the two-block system between the initial state and the equilibrium state? (c) What is the change in the entropy of the two-block system? The heat capacities of copper and lead are 386 J/kg·K and 128 J/kg·K, respectively.

Answer 1

1) T(final) =238.7K

2) Change in internal energy = 0

3) ΔS = +1.24 J/K

Step-by-step explanation:

a) Heat lost by Cu (copper) block = Heat gained by lead (Pb)

`-q(Cu) = q(Pb)`

`-[m(Cu)*c(Cu)*(T2-T1,Cu)] = [m(Pb)*c(Pb)*(T2-T1,Pb)]`

where m = mass of the substance

c = heat capacity

T2 and T1 are the final and initial temperatures

Substituting the appropriate values:

`-[0.032*386*(T2-310)] = [0.0887*128*(T2-161)] \\\\= -[12.352T2-3829.12] = 11.354T2-1827.93\\\\T2 =238.7K`

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:

`\Delta S = \int (dq)/(T)= \int (mc*dT)/(T)= mc*ln((T2)/(T1))`

For the 2 block system:

`\Delta S = \Delta S(Cu) + \Delta S(Pb)= m(Cu)c(Cu)*ln((T2)/(T1))+m(Pb)c(Pb)*ln((T2)/(T1))`

`= 0.032*386*ln(238.7)/(310) +0.0887*128*ln(238.7)/(161) \\\\=-3.228 + 4.471 = +1.24 J/K`