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problem 1
SATELLITE 

Given:
Consider a satellite containing electronics electronics that consume power, P . All of this power is turned into heat that must be eliminated, somehow.
Find: You've learned that heat can be transported by (1) conduction, (2) convection, or (3) radiation. Which process is important here? Answer 1, 2, or 3.
Solution: 3

A

Given: Suppose the satellite is a spherical black body with a radius r at room temperature T
Find: The power is radiated by this object
Solution:
P = JA = (sigmaSB)(T^4)(4pi*r^2)

B 
Given: The satellite's electronics generates x Watts
Find: The equilibrium temperature, Teq , of the satellite
Solution:
P = JA
x Watts = (sigmaSB)(T^4)(4pi*r^2)

C 

problem 2
HEAT FLOW, ENTROPY, AND FREE ENERGY 

Given:
As you know, equilibrium is reached when T1 = T2 , which corresponds to a maximum in entropy, or minimum in free energy. Consider two identical cups of water, initially at temperatures of T1°C and T2 °C. The two cups are brought into thermal contact. In order for them to reach equilibrium at Teq °C, x J flows from the hot cup to the cold one. Assume that the heat capacity C is constant.
Find: Calculate the C of each cup.
Solution:
dQ = Cdt, integrate to get x = CT, between T1 and Teq.

A

Given: The same thing
Find: The entropy change of the cold cup as it is heated to Teq °C
Solution:
deltaS = C*ln(Teq/T1). Make sure your temps are in Kelvin.

B 
Given: Same
Find: How much the total entropy of the two cups change as they equilibrate to Teq °C ?
Solution:
deltaStotal = C*ln(Teq/T1) + C*ln(Teq/T2)

C 
Given: Same
Find: How much the total free energy of the two cups, referenced to the equilibration temperature of 33°C, changed during this process
Solution:
deltaF = deltaU  Teq*deltaS, deltaU = 0

D 

problem 3
CALCULATING ENTROPY CHANGES 

Given: Questions a) and b) refer to this situation:
A macroscopic box is divided into two halves by a partition. Each half contains N gas molecules at the same pressure and temperature. Consider what happens to the entropy when the partition is removed.
Find: If the two halves contain the same gas ( e.g. , both are oxygen), the entropy will
 decrease significantly.
 remain almost the same (ie, a reversible process).
 increase significantly.
Solution: 2

A

Given: Same
Find: If the two halves contain different gases ( e.g. , oxygen and nitrogen), the entropy will
 decrease significantly.
 remain almost the same (ie, a reversible process).
 increase significantly.
Solution: 3

B 
Given: Part b is an example of entropy change caused by removal of a constraint (the partition). Now consider the entropy change during quasistatic processes, ones in which V, p, and T change very slowly and are always well defined.
Start with V1 liter of argon (an ideal, monatomic gas) at pressure, p = p1 kPa, and temperature, T = T1 K.
Find: Calculate the entropy change ?S if we were to let the gas expand to V2 liters (keeping the temperature constant).
Solution:
Isothermal means no change in temp, which means no change in T. Thus Q = Wby.
?S = Q/T = Nk*ln(V2/V1). You can get Nk from Nk=pV/T

C 

Given: Start with n moles of argon (an ideal, monatomic gas) at pressure, p = p kPa, and temperature, T K.
Find: The entropy change ?S if we were to add x J of heat to the gas (keeping the volume constant).
Solution:
deltaU = (3/2) nR(TfT)
solve for Tf
?S = nR*ln(V2/V1) + (alpha)nR*ln(T2/T), nR*ln(V2/V1)=0

D 

problem 4
ADIABATIC PROCESS 

Given:
For quasistatic processes, the entropy of an ideal gas changes in the following way:
?S = nR ln(V f /V i ) + anR ln(T f /T i ). In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure p1 kPa and temperature T1 K, is allowed to expand adiabatically from V1 m^3 to V2 m^3 while doing work on a piston.
Find: The change in entropy due to the volume change alone, ignoring any effects of changing internal energy
Solution:
?S = nR*ln(V2/V1), nR = pV/T

A

Given: same
Find: For this adiabatic expansion, the final temperature
Solution:
T1V1^alpha = T2V2^alpha
alpha = 3/2
solve for T2

B 
Given: Same
Find: The change in entropy solely due to the change of temperature, ignoring the entropy change due to the volume change
Solution: ?S = (alpha)nR*ln(T2/T1)

C 
Given: Same
Find: The total change of the entropy in this adiabatic expansion
Solution:
?S = (part A) + (part C) = 0


