(a) Find the equilibrium temperature that results when one...

(a) Find the equilibrium temperature that results when one kilogram of liquid water at $373 \mathrm{~K}$ is added to two kilograms of liquid water at $283 \mathrm{~K}$ in a perfectly insulated container. (b) When heat is added to or removed from a solid or liquid of mass $m$ and specific heat capacity $c,$ the change in entropy can be shown to be $\Delta S=m c \ln \left(T_{f} / T_{i}\right)$ where $T_{\mathrm{i}}$ and $T_{\mathrm{f}}$ are the initial and final Kelvin temperatures. Use this equation to calculate the entropy change for each amount of water. Then combine the two entropy changes algebraically to obtain the total entropy change of the universe. Note that the process is irreversible, so the total entropy change of the universe is greater than zero. Assuming that the coldest reservoir at hand has a temperature of $273 \mathrm{~K},$ determine the amount of energy that becomes unavailable for doing work because of the irreversible process.

Key Concepts

Unavailable Energy and Exergy Loss

Not all energy in a system is available for doing work; some portion becomes unavailable due to entropy production during irreversible processes. The concept of exergy loss accounts for the energy that is degraded, meaning that when heat transfers occur—especially in processes with reservoirs at different temperatures—a portion of that energy is lost to the surroundings as entropy increases, reducing the overall efficiency of energy utilization.

Irreversible Processes and the Second Law of Thermodynamics

Irreversible processes are characterized by an overall increase in the entropy of the universe, as stated by the second law of thermodynamics. In mixing problems, even though the system reaches a new equilibrium state, the process is irreversible because the combined entropy change (of the system and surroundings) is positive, indicating a net loss in the potential to do work.

Entropy Change in Temperature Variations

Entropy quantifies the degree of randomness or disorder in a system and typically increases during irreversible processes. For a substance undergoing a temperature change with a constant specific heat capacity, the change in entropy can be calculated using the formula ?S = m c ln(Tf/Ti). This formula helps quantify the increase in disorder when portions of water at different temperatures come to a common equilibrium.

Specific Heat Capacity

Specific heat capacity measures the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Kelvin. It plays a critical role in thermal calculations such as determining how much energy is exchanged when water of one temperature is mixed with water of another, by relating heat transfer to temperature change.

Thermal Energy Conservation and Equilibrium

This concept involves understanding that in an insulated system, no heat is lost to the surroundings so that all the energy exchanged between components must balance. When substances at different temperatures are mixed, the final equilibrium temperature is determined by setting the heat lost by the warmer substance equal to the heat gained by the cooler substance, ensuring conservation of energy.

Transcript

00:01 In this problem, we're going to be looking at the mixing together of two different amounts of water at two different temperatures.

00:09 And in the end, then look at the entropy aspect of this process.

00:13 So step a, though, the first thing we need to do is find the equilibrium temperature.

00:18 And so what we're going to do is notice that m2 is lesser in temperature than m1.

00:23 So m2 is going to gain heat, and one is going to lose heat.

00:27 So on the left, i'm going to write what is actually gained by m2.

00:32 So that's m2.

00:36 Cw, the specific heat of water, t final, minus t2, equal minus the heat lost by m1.

00:51 M1, cw, t final, minus t.

00:57 Now, before i go on, let me make a couple of comments here.

01:01 You might say where this minus sign coming from.

01:03 Well, many places you will see that they'll say when there's going to be a loss, just switch the t's.

01:09 That's fine.

01:10 That works fine.

01:11 There's also another way.

01:13 Realizing that this is a closed system, the net then is equal to zero.

01:20 No heat is entering the system or leaving it.

01:22 So everything is going on internally.

01:24 So that would be equal to the heat gain by one part plus the heat lost by the other.

01:31 Other that's equal to zero it all comes out the same it's whatever you're comfortable with so now let's do a little algebra that's factor out the t tf found temperature i have an m2 and should mention c w is canceled so we don't have to worry about those we have m2 we have a minus m1 on the right so that becomes plus m1 on the left and we have a minus m2 t2 so that's going to become m2 and t2 on the right.

02:07 And we got minus m1, minus t1.

02:09 So that's going to be plus m1, t1.

02:14 So to get t final, we just got to divide by a total mass.

02:18 So it's pretty straightforward.

02:19 We can put in our numbers now.

02:21 So this will be 2 kilograms, 283 kelvin, plus 1 kilogram, 373 kelvin, over two kilograms plus one kilogram.

02:52 And this works out to give us an equilibrium temperature of 313 calvert.

03:02 And that's good, because you've got to be between 283 and 273...