You probably know that when two objects at different temperatures are brought in

Question

You probably know that when two objects at different temperatures are brought into contact, heat always travels from the hotter object to the colder object; never the reverse.

Imagine that you have two gold blocks, one with a mass 289 g at a temperature of 20.4 oC, and another with a mass 395 g at a temperature of 117 oC. The two blocks are brought into contact with each other in an insulated container.

Specific heat of gold = 0.129 JK-1g-1

What is the sign of the entropy change for the heat transfer (enter positive or negative) that takes place?

Sign of entropy change = positive

What is the final temperature of the two gold blocks?

Final temperature = 76.2 oC

What is the entropy change for the heat transfer described in this question?

Entropy change =

Explanation / Answer

m1=289g T1=20.4C=293.4K

m2=395g T2= 117C = 390K

At equilibrium, the two blocks of gold will have the same temperature .

Energy is conserved during process.

heat lost by hotter block = heat gain by cooler block

delta q= m s deltaT

let at equlibirium Temp =T

395 xs (390-T) =-289 xs x (T-293.4)

1.36678(390-T) = T-293.4

533.0442 -1.36678T = T- 293.4

826.444 = 2.36678T

T= 349.18=76.185

Tds equation,
Tds = dh - vdp, dp~0(incompressible)
ds = (cp/T)dT
S = m*C*ln(T/Ti)

Entropy change 1
S1 = m1*C*ln(T/T1) = 289x0.129 ln 349.18 /293.4=289x0.129 x0.173953=6.485J/K

Entropy change 2
S2 = m2*C*ln(T/T2) = 395x0.129 ln(349.18/390)=-5.6337

Total entropy change of heat transfer,
S = S1 + S2 =6.485-5.6337= 0.8513J/K

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