T is not constant through the warming. Obtain a differential equation from ΔS=ΔQ/T and solve it.Art_Vandelay said:Homework Statement
Imagine that the temperature of 255 g of aluminum sitting in the sum increases from 278 K to 294 K. By how much has its entropy increased?
Homework Equations
Q=mcΔT
ΔS=Q/T
The Attempt at a Solution
Q=(255 g)(.90 J/gK)(294 K -278 K)
Q=3672 J
ΔS=Q/T
ΔS=3672 J / 294 K
ΔS= 12.5 J/K
This answer is incorrect, but I don't know why.
haruspex said:T is not constant through the warming. Obtain a differential equation from ΔS=ΔQ/T and solve it.
No, I mean consider just a very small change.Art_Vandelay said:Would a differential equation be subtracting ΔSCOLD from ΔSHOT?
Hmm... Would I use 1/T = dS/dU? But how could I figure out dU without a given pressure?haruspex said:No, I mean consider just a very small change.
If a little heat, dQ, enters the metal, there is a corresponding small change dT to the temperature and a small gain in entropy, dS = dQ/T.
You need a second equation to relate dT to dQ.
It would almost be equal to zero?TSny said:Following haruspex's lead, you need an expression for dQ to use in dS = dQ/T.
What does the equation Q = mcΔT look like for an infinitesimal amount of heat?
Well, yes, but TSny means using dQ, dT. The Q in Q = mcΔT is a change in heat content, not the total heat content. So really it should ΔQ = mcΔT. But, conventionally, Δ is used for (possibly) substantive differences, whereas d is used for infinitesimal changes.Art_Vandelay said:It would almost be equal to zero?
Oh! Ok, so I would end up with S*T=mc(dT). Would the temperatures then negate each other, or would I need to divide mc by T?haruspex said:Well, yes, but TSny means using dQ, dT. The Q in Q = mcΔT is a change in heat content, not the total heat content. So really it should ΔQ = mcΔT. But, conventionally, Δ is used for (possibly) substantive differences, whereas d is used for infinitesimal changes.
So, rewrite Q = mcΔT using dQ, dT, and substitute in the entropy equation to eliminate dQ.
Again, that's not quite right. It's an infinitesimal change in S, so it's dS*T=mc*dT.Art_Vandelay said:Oh! Ok, so I would end up with S*T=mc(dT). Would the temperatures then negate each other, or would I need to divide mc by T?
Unfortunately, I've never used a differential equation before. Is it calculus-based?haruspex said:Again, that's not quite right. It's an infinitesimal change in S, so it's dS*T=mc*dT.
Can you solve that differential equation?
Art_Vandelay said:Unfortunately, I've never used a differential equation before. Is it calculus-based?
Here's my attempt:
dS*T=mc*dT
dS=((.255 kg)(910 J/kg K)(>0K?)/294 K
dS=.79 J/K
Since dS is infinitesimal, would ΔS be found by multiplying by ΔT?TSny said:Sorry, but that's not right at all.
Solving dS*T = mc*dT does involve calculus. Suppose you rearrange the equation as dS = mc*dT/T. In this form the variables S and T are "separated", with S on the left and all T's on the right. Can you see what to do to obtain ΔS from dS?
Art_Vandelay said:Since dS is infinitesimal, would ΔS be found by multiplying by ΔT?
I wish I did have some knowledge of calculus, but I do not. I really have no clue how to evaluate the practice problem.TSny said:No.
Do you have some calculus background?
As a little exercise that is relevant to this problem, can you evaluate $$\int_{x_1}^{x_2} dx$$ in terms of ##x_1## and ##x_2##?
How about this one? $$\int_{x_1}^{x_2} \frac{dx}{x}$$
TSny said:This problem requires calculus to solve it. If this is a problem from a course in physics that does not use calculus, then I don't see how you would be expected to solve it.
Could it be that the instructor or textbook has given you a formula (without derivation) that you can use for this particular type of problem? The formula would contain a logarithm function.
[Edit: You can get an approximate answer by using ΔS = Q/Tavg, where Tavg is an average temperature of the aluminum during the heating process.]
Entropy increase of aluminum refers to the natural tendency of aluminum to become more disordered and less organized over time. This is due to the fact that the atoms in aluminum are constantly moving and rearranging themselves, leading to an increase in randomness and disorder.
Entropy increase in aluminum is driven by the second law of thermodynamics, which states that the total entropy (or disorder) of a closed system will always increase over time. Since aluminum is a closed system, its entropy will naturally increase as the atoms move and rearrange themselves.
Temperature plays a significant role in the entropy increase of aluminum. As the temperature increases, the atoms in aluminum gain more energy and move more rapidly. This leads to an increase in disorder and therefore, an increase in entropy.
No, entropy increase of aluminum cannot be reversed. The second law of thermodynamics states that entropy will always increase in a closed system, and therefore, it is not possible to reverse the natural tendency of aluminum to become more disordered over time.
The increase in entropy of aluminum can have various effects on its properties. It can lead to changes in its physical structure, such as increased malleability and decreased strength. It can also affect its chemical properties, making it more reactive and prone to corrosion. Additionally, the increase in entropy can impact the thermal and electrical conductivity of aluminum.