A ceramic teapot (e = 0.70) and a shiny metal one (e = 0.10) e...

Question

A ceramic teapot (e = 0.70) and a shiny metal one (e = 0.10) each hold 0.55 L of tea at 85°C. (a) Estimate the rate of heat loss from each, and (b) estimate the temperature drop after 30 min for each. Consider only radiation, and assume the surroundings are at 20°C.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the Stefan Boltzmann law. This problem says you are a barista in a cafe and have brewed two spherical pots of coffee with 0.75 L in each at the initial temperature of 90 °C. One pot is made of glass with an E miss E is equal to 0.85 and the other is made of dull minum. E is equal to 0.15. The surrounding temperature is 25 °C. There are two parts to this question. For part one estimate the rate of heat loss from each pot due to radiation. And for part two, calculate the temperature drop from each pot after 45 minutes considering only radiative heat loss. And we're giving a hint that the specific heat capacity of water C is equal to 4.18 joules per gram degrees Celsius. Now, for part one, we need to calculate the heat loss of each pot due to radiation. So here we're going to use our stuff in Boltzmann Law. So if we call for that, we'll have delta Q divided by delta little T is equal to E multiplied by sigma multiplied by a multiplied by the quantity of capital T raised to the fourth minus capital TS raised to the fourth where delta Q divided by delta little T is our rate of heat loss. E is our emissivity sigma is the stuff on both constant A is the surface area of our pot. Capital T is the temperature of the pot and capital T SS is the temperature of the surroundings. Now, if I look at the information I'm given the problem, I was actually given the volume of the spherical coffee pots, but not the surface area. But we can actually relate the volume to the radius and then the radius to the area. Therefore connecting the volume to the area. So that's what we're going to do now. So recall your formula for a volume of a sphere that could be V is equal to four divided by three multiplied by pi multiplied by R cubed. So I need to actually solve this for R. When I do that, I'll get R is equal to the quantity of three multiplied by V divided by four pi. And that's in quantity and that's what we raise to the one third power. Do I need to use that in my area formula? So recall your formula for the surface area of the sphere that's gonna be A is equal to or pi R squared. So plug in our expression for R, we're going to get A is equal to or pi multiplying the quantity of three V divided by or pi in quantity raised to the two thirds power. So we now take that formula for area and plug it back into our step. And Boltzmann equation will have delta Q divided by delta little T is equal to E multiplied by sigma. And now for area we'll plug in our expression. So that's gonna be multiplying the quantity of or pi multiplied by the quantity of three V divided by or pi in quantity raised to the two thirds. Then that quantity is gonna be multiplied by the quantity of capital T to the fourth minus capital TS to the fourth. Now, the only other thing we need to worry about is our temperatures. So the temperatures in our step and Boltzmann equation must be in Kelvin and here we were given degrees Celsius. So we'll need to go ahead and convert those temperatures. So for the temperature of the pots, that's gonna be capital T and that's gonna be equal to. And here we're gonna convert the 90 °C into Kelvin. And to do that, I'm gonna take the 90 add 273 to it. Now give me my temperature in Kelvin. That means that capital T is equal to 363. Kelvin. We'll do the same thing for the temperature for the surroundings. That'll be capital Ts is equal to and here, its temperature is 25 °C. So I'll have 25 plus 273. That'll give me my temperature in Kelvin. And so that means that's gonna equal 298. Kelvin did not have both temperatures in Kelvin. I can actually plug in all of our values into our formula for the rate of heat loss. Now, so I have to do two different calculations and our volume and temperatures are going to be the same for both parts for both pots. What we're going to want to do is actually calculate value for everything except for e our emissivity. So we'll leave it as a variable that would just have our emissivity multiplied by a number. And it'll make calculating the different um rates of heat loss for the two different pots a lot easier. They're doing that. Coming back to our Stefan Boltzmann law will have delta Q divided by delta level T is equal to e multiplied by our Stefan Boltzmann constant, which is 5.67 multiplied by 10 to the negative eight that has units of watts per meter squared. Kelvin to the fourth. Then that's gonna be multiplied by the quantity of four pi multiplied by the quantity of three multiplied by the volume, which is 0.75 L which we need to convert into meters cubed. And to do that, we call it to convert from liters into meters cubed. You need to multiply by 10 to the negative three. So we'll have 0.75 multiplied by 10 to the negative three meters cubed. And then we're going to divide that by four pi and then that quantity is raised to the two thirds power and then that quantity is multiplied by the quantity of capital T to the fourth which is 363. Kelvin, that's gonna be raised to the fourth power minus capital Ts to the fourth, which is 298 Kelvin race to the fourth. So plugging all those values into our calculator will get delta Q divided by delta T is equal to e multiplied by 21.45. What? So here, I've just rounded our value to four significant figures. So now we can use this expression to calculate the rate of heat loss for each of our pots. So we're actually gonna start off with the glass pot. We will call that rate of heat loss delta QG divided by delta little T and that's gonna be equal to the emissivity of the glass pot which is the 0.85 and that's multiplied by the 21.45. What? So plug in those values in my calculator, I'll get delta qg divided by delta T is equal to 18.2 watts. You have just kept three significant figures. But to match the number of significant figures that were given in the problem, we need to actually round this to two significant figures. So rounding that to two significant figures, I'll just have 18 watts. So that's one of the answers I was looking for for part one, we'll do the same thing for the aluminum pot. So we'll have delta QA A for the aluminum divided by delta Little T is equal to its Emmis is 0.15. It's going you multiply again by the 0.1 0.45. What? And so plug in those values into my calculator, I'll get delta Q A divided by delta little T is equal to 3.22 watts. Here again, I've kept three significant figures but I really should round it to two. So this is gonna be approximately equal to 3.2 watts. That could be the other half of my answer for part one. Now, for part two, I need to actually calculate the change in temperature that has occurred due to the radiation heat loss um over 45 minutes. So for part two, we first need to recall the amount of heat that's lost in order to change a temperature. So recall for that formula, you're gonna have Q is equal to D multiplied by M multiplied by delta T Q is the heat that's been lost or gained in this case. But in this case, it's gonna be lost is our specific heat capacity. M is going to be the mass of the water in the coffee pot. And delta T is going to be the change in temperature. We need to actually solve this for the change in temperature. So for delta T, we get delta T is equal to two divided by DM. Now, we don't have the amount of heat that we have lost directly, but we do have the rate at which heat is lost. So I'm going to do is multiply the right hand side by one in the form of delta little T divided by delta little T. So when I do that, I'll get delta capital T is equal to one divided by cm, modify delta Q divided by delta little T multiplied by delta little T. So now we can actually plug in values. We were given the uh specific heat capacity in the problem. That was the 4.18 joules per gram degree Celsius. We just calculated our delta Q divided by delta little T for each of the pots. And we are also given our delta T, which is the 45 minutes, but we weren't given the mass. And the problem, however, we weren't given the volume. And since we're dealing with water here, which is what in the coffee pot, um we can actually use the density of water to calculate the mass. So we call for the mass that's going to be M is equal to our density row multiplied by our volume B. So M is going to be equal to I can look up the density of water that is 1000 grams per liter. I don't want to multiply that by our volume, which is 0.75 L. That means that the mass M is gonna be equal to 750 g. Do not have all the values that need to calculate my change in temperature. So we're gonna do this for each of our pots. We're gonna start off with the glass pot. And so we'll have delta capital TG for the glass. That's gonna be equal to one divided by the quantity or our specific key capacity. That is the 4.18 jules burr gram degree Celsius. This could be multiplied by our mass, which is the 750 g. And then that is going to be multiplied by our rate of heat loss, which is the 18.2 watts. And this could be multiplied by our time, which is 45 minutes. But I'm actually gonna need to convert that minutes into seconds because if you recall a watt is a jewel per second. So to cancel out that second, I need to have my time here also in seconds. So we'll multiply our 45 minutes by the conversion factor of 60 seconds divided by one minute. So I can now plug all those values in my calculator and I'll get delta capital TG is equal to 16 °C. And here I've rounded to two significant figures that's one half of our answer for part two. And so for the other half, we'll do the same thing. But for the aluminum pot, we will have our delta capital T A is equal to one divided by the quantity of the 4.18 jules per gram degrees Celsius multiplied by the 750 rams. Then that's gonna be am pride by our rate of heat loss, which in this case is the 3.22 watts. This could be again, multiplied by 45 minutes. And again, I'm going to need to convert that into seconds by multiplying by 60 seconds divided by one minute. So plugging those values to my calculator, I'll get delta, the capital T A is equal to 2.8 °C. And here again, I've already rounded to two significant figures. So those are gonna be my answers for part two. So just recap of what we've done here. For part one, we use our E and Baltimore law to calculate the rate of heat loss for each of our pot pots. For part two, we use our formula for the amount of heat that's required to rate the temperature of a substance. But here we replace our heat here by our heat rate multiplied by our time that would give us the total heat. And then we just plug in values to calculate the change in temperature for each of our pots. So I hope that this has been useful and I'll see you in the next video.

A ceramic teapot (e = 0.70) and a shiny metal one (e = 0.10) each hold 0.55 L of tea at 85°C. (a) Estimate the rate of heat loss from each, and (b) estimate the temperature drop after 30 min for each. Consider only radiation, and assume the surroundings are at 20°C.

Key Concepts

Emissivity

Stefan-Boltzmann Law

Heat Transfer

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