(II) A 110-g insulated aluminum cup at 35°C is filled with 150...

Question

(II) A 110-g insulated aluminum cup at 35°C is filled with 150 g of water at 45°C. After a few minutes, equilibrium is reached. Determine the total change in entropy as a result of the mixing process (use ∆S = ∫ dQ / T ).

Accepted Answer

Welcome back. Everyone in this problem, a steel cup with a mass of 200 g initially at room temperature of 20 °C is filled with hot tea weighing 0.6 kg. The tea is initially at its boiling point temperature of 100 °C. After some time, thermal equilibrium is reached at a temperature of 97 °C using or changing entropy being equal to the integral of one divided by T with respect to Q compute the entropy change due to this process assume that the specific heat capacities for steel and water are 490 joules per kilogram per Kelvin and 4180 joules per kilogram per Kelvin respectively. A says that the entropy change is negative 0.62 joules per Kelvin B 2.6 joules per Kelvin C 3.2 Joules per Kelvin and D 5.4 joules per Kelvin. Now, if we're going to figure out the entropy change, we're given quite a bit of information here. So let's try to make note of what we know. So we know that our tea is poured into our cup. What do we know first about the cup. Well, we know for starters that its mass is 200 g or 0.2 kg. We know it's initially at our room temperature ts of 20 °C. And if we write that in Kelvin by adding 273.15 that makes it 293.15. Kelvin. OK. And we also know that for our cup, its heat, its specific heat capacity. So let's call that CS our steel is 490 joules per kilogram per Kelvin. So that's what we know about our steel cup. What about our tea? Well, we know that the mass of the T is 0.6 kg. OK. We also know that the tea is initially at its boiling temperature. So the T temperature of T is 100 °C or 373.15 Kelvin. OK. And we also know that the specific capacity for our team is equal to 4180 jules per kilogram. Yeah. And finally, we know that the final equilibrium temperature TF is equal to 97 °C or 370.15 Kelvin. OK. And we want to use all of this to figure out our entropy change. Now, how can we relate this information to our entropy change? Well, we know that our entropy change, OK. Delta S is going to be equal to the entropy change. For our steel cup plus the entropy change for our T. Now, in our problem, we were told that or we were given a formula for our entropy change. We know that the entropy change delta S is equal to the integral of DQ divided by our temperature T OK. So using this formula then that means our total entropy change is going to be equal to the integral of DQs, which is going to be the derivative of the heat change for our steel cup divided by T OK, between the boundaries of the temperature for the steel cup TS and the final temperature. TF And now we're gonna add that to the integral of DQT divided by T. So DQT is the change in temperature for our T sorry or the change in heat for RT OK, between the temperature for RT are between the bonds for the temperature of for our T and our final temperature. Now here. So we have an expression but the issue is we don't know yet or our problem didn't give us what DQ is equal to. However, we can recall, OK, that DQ that change in heat, that derivative of our heat change is going to be equal to the mass multiplied by the specific heat capacity multiplied by the derivative of the temperature. OK. So if we use this formula here, if we use that formula in our problem, we can express DQ in terms of T. Thus, we can integrate and then solve for our entropy change. So by doing that, OK, then we can rewrite this as MC delta T, OK. Oh Sorry, not the, the D MC. And let me write a proper subscript here. So that's MS CS DT divided by T, OK? Between TS and TF OK. Plus integral between TT and TF of MTCT delta T divided by T OK. And now we can take out our contents. OK. And thus, we're gonna get delta S, once we do that to be equal to MS CS multiplied by the integral between TFTs of DT divided by T plus NTC multiplied by the integral between TT, OK. And the TF of DT divided by T. So we have our integrals. The difference is just our boundaries. So now we can go ahead and integrate, OK. I know when we do that, I think I can make, write it beside it. When we do that, then we're gonna get MS CS multiplied by the natural log of TF divided by TS, OK. Because that would be the integral of DT over T or in other words, one divided by T with respect to T, OK. And then we're gonna be adding this to MTCT multiplied by the natural log of TF divided by TT. Now we can go ahead and substitute our values here because we know what all of these values are. So now that tells us then that the change in entropy then by substituting is going to be equal to the mass for ST C 0.2 kg. Multiplied by its specific heat capacity. 490 joules per kilogram per Kelvin. OK. Multiplied by the natural log of our final temperature. 370.15 Kelvin divided by the initial temperature for our steel cup 293.15 Kelvin. On the other hand, when it comes to our tea, then we're going to be multiplying its mass of 0.6 kg by its specific heat capacity of 4180 joules per kilogram per Kelvin. OK. Multiplied by the natural log of the final temperature. 370.15 Kelvin divided by the initial temperature of our boiling water. 373.15 Kelvin. So now we can go ahead and solve this expression here, put this into our calculator to solve for the change in entropy. Now, when we do that, then we should find that we get the change in entropy to be equal to 2.611 Joules per Kelvin. Or if we write that to two significant figures like the rest of our answers, approximately 2.6 joules per Kelvin. When we look back on our answer choices, then we can see that B is the correct answer. Thanks for watching everyone. I hope this video helped

(II) A 110-g insulated aluminum cup at 35°C is filled with 150 g of water at 45°C. After a few minutes, equilibrium is reached. Determine the total change in entropy as a result of the mixing process (use ∆S = ∫ dQ / T ).

Key Concepts

Entropy

Heat Transfer

Thermal Equilibrium

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