Question

A sample of 1.00 mol of silver at 150 degrees celsius is placed in contact with 1.00 mol of silver at 0 degrees celsius. Calculate (a) the final temperature of both silver samples; (b) the delta S for the hot Ag sample; (c) the delta S for the cold Ag sample; and (d) the total delta S ofthe system. (e) Is the process spontaneous? How do you know? Assume a constant heat capacity for Ag of 25.75 J/mol-K.

Answer 1

a)

final temperature for

Qcool = -Qhot

n1*Cp*(Tf-Ti) = n2*Cp*(Tf-Ti)

for

1*Cp*(Tf-150) = -1*Cp*(Tf-0)

cancel common terms

Tf-150 = -Tf-0

2Tf = 150

Tf = 75 °C

b)

dS for hot sample:

dS = n*Cp*ln(Tf/Ti)

MW f Ag = 107.86820 g/mol

Cp Ag = 0.240 Jg/c

Tf = 75°C = 75+273 = 348 K

Ti = 150°C = 150+273 = 423K

dS = 25.75*ln(348 /423) = -5.0526 J/K

c)

dS for Silver cool

Cp Ag = 0.240 Jg/c

Tf = 75°C = 75+273 = 348 K

Ti = 0°C = 0+273 = 0 K

dS = 1*25.75*ln(348 /273) = 6.250J/K

d)

dSsystem = dScool + dShot = 6.2839 -5.0526 = 1.2313 J/K

e)

this must be spontanoeus, since it follows all the laws of thermodynamics

Tlow increases and Thigh decreases, this is zeroth law

dS universe increases, also according to forward direction