Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m^2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

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Accepted Answer

Hey everyone in this problem, we have a 0.4 m long aluminum rod joined end to end with a 0.75 m copper rod for a heat transfer experiment, All sections on the compound Rod have a cross sectional area of 0.0086 m squared. The copper end is inserted in boiling water while the aluminum end is inserted in an ice water mixture. The experiment is conducted at atmospheric pressure. The Rod is wrapped to prevent heat loss to the surroundings, and we are asked to determine the mass of ice that will melt in 10 minutes using the heat supplied by the Rod. So let's just draw a little diagram of what's going on. Okay, so we can imagine that we have this pot or container of boiling water. Okay, now we're atmospheric pressure. So if we have water that's boiling then we're gonna be at 100°C. And we have a copper rod inserted into this boiling water because we have the end of the copper rod in this boiling water. So this is copper. And we know that the length of this rod is 0.75 m now joined end to end with this copper rod is an aluminum rod. Okay, and it is placed in an ice water mixture. So we have this container of ice water. Okay, now, when we have an ice water mixture, that means we're right at the boundary of Water freezing. Okay, so we're going to be at 0°C here this rod is aluminum, just gonna short form a loon. Okay, and we know that this has a length of 0.4 m given in the problem. Alright, so this is kind of gives us an idea of what's going on. Now. This is a heat transfer experiment and we know that heat is going to go flow from high temperature to low temperature. Okay, so in this problem our heat flow is going to the right okay, through that copper, from the boiling water, through the copper rod, through the aluminum rod over to the ice water, which is gonna cause some of that ice to melt. Okay. Alright, so we are asked to find the massive ice that will melt using the heat supplied. Okay, now, when we're talking about the heat here and the ice melting. Okay, let's recall that the heat Q. Is going to be equal to m times of latent heat of fusion LF. Okay, in the case where we're melting. Alright, so we know LF. The latent heat of fusion that is a constant that we can find. So if we can figure out Q. Then we know that we can find the mass M by Q divided by L. F. The latent heat of fusion. Alright, well this is a heat transfer problem. Let's recall that our heat current age Is equal to Q. The Heat divided by T. The time. Okay, now we know the time we're told 10 minutes. So if we can figure out what H is the heat current? This will allow us to calculate Q. And then we can use that Q. To calculate the mass. M. Hm. Alright, so let's look at our heat current. Okay let's start with copper. Okay. And the heat current H And we're gonna call it H subscript C. For copper, it's going to be equal to K. And again K. Sub C. For copper. Where K. Is the thermal conductivity of the material of the rod? Okay. A cross sectional area th minus T. C. K. The temperature of the hot end minus the temperature of the cold end of the rod divided by L. C. The length of our copper. Now Casey that thermal conductivity this is a constant. We can look up in a table in our textbook or that your professor provided, it's gonna be 385. What? Per meter? Degrees Celsius. And this you can use in degrees Celsius or Calvin. Ok. In this problem we're talking about degrees Celsius 100 0. so we'll do it in degrees Celsius but you can do it in Calvin as well. Now the cross sectional area and let me just move this H equals Q over T. Equation up here. So it's out of the way. So again that heat current is going to be equal to the quantity of heat Q divided by the time. T. Alright so back to our heat current for copper, we have the cross sectional area. 0.86 m squared. Okay now the temperature at the hot end of our copper rod. Well that's the end that's in the boiling water. So that's going to be 100°C. Okay, now the copper or the temperature at the cold end? Well, we don't know. Okay, that's going to be this end here, which is going to be this boundary point. Okay, so let's just call it T. And we know the length of our copper Rod is 0.75 m. Alright, so we can't calculate H because we don't know t let's move over to the aluminum rod and see what we can find. And again, H. A. Okay, current for aluminum is going to be K thermal conductivity and in this case of aluminum times a the cross sectional area times th minus T. C. Again divided by L. A. The length of the aluminum run. Alright, so, okay, the thermal conductivity for aluminum, we can look this up 205 watt per meter degrees Celsius times 0.86 m squared. Okay, we're told that the road has the same cross sectional area for both sections. Now our temperature and the hot end, this is actually going to be the temperature at the boundary. Okay, Because we know that the heat flow is going from left to right. We know that that right hand end in that ice water is gonna be the coldest temperature we have. Okay, and so the temperature at the hot end of the aluminum rod is going to be t It's gonna be the same temperature at the cold end of the copper rod. The temperature of the boundary minus zero degrees Celsius. And again this temperature t here corresponds to the temperature at the boundary of the two rods. Alright. And then we have to divide by the length and the length of our aluminum rod is 0. m. Okay, so we're trying to calculate h we can't do that until we can find this temperature t at the boundary. Okay. No, let's think about this. Okay, because we have the rod wrapped, we have no heat loss to the surroundings. Okay, So if we have no heat loss. Two surroundings when we go from the copper rod to the aluminum rod, this tells us that the heat current is the same for both sections. Hm. So we have no heat loss. That means we have no energy lost. And the heat current is going to be the same for copper and aluminum. Well, if the heat currents the same, we can set them equal. Okay, so we have the heat current for copper is equal to the heat current for aluminum and we can set them equal to each other. So we get 385 watts per meter degrees Celsius time 0. m squared times 100 degrees Celsius minus T divided by 0. m is equal to watts per meter degrees Celsius. 0.86 m squared times t -0°C, All divided by 0.4 m. Now, we have an equation that has just one unknown this value of t. The temperature of the boundary. Okay? So we can go ahead and solve for that temperature at the boundary. That's going to allow us to solve for H. Okay, Alright, So first thing we can do, we can divide by 0.86 m squared because it's in both sides. That term is going to divide out. Now let's multiply up by 0.4 and 0.75 respectively. On each side we get 0.4 m times. And then we're going to expand the 385 times 100 degrees Celsius minus T. So 385 watts per meter degrees Celsius times 100 degrees Celsius. This is going to give us 38,500 watts per meter. And then we have minus watts per meter degrees Celsius times that temperature T. Okay. And on the right hand side we get 0.75 m. And then we're gonna multiply here. So we get 205 watts per meter degrees Celsius times the temperature T. All right, so let's move this term with the T over to the right hand side. Okay, on the left hand side, we're gonna be left with 0.4 m times 38,500 watts per meter. That's gonna give us 100 or sorry, 15,400 watts. And on the right hand side, we factor out that T. Okay, we get 0.75 times 205. We got 153.75. Okay, watt per meter degrees Celsius times meter. So that's gonna be what per degree Celsius? And then we add this term from the right hand that we brought over from the left hand side. 0.4 m, times 385 watt per meter degrees Celsius. Which is gonna give us 100 and 54 watts per degree Celsius. And so t if we divide is gonna be equal to 15, 400 watts, divided by 307.75 watts per degrees Celsius. This is going to give us a unit per of degrees Celsius, which is what we want for temperature. And we get 50.4 degrees Celsius. Okay, Alright, so now we found the temperature at the boundary T. Okay, so that temperature at the boundary of those rods is 50.04°C. And in this case it's really close to the kind of the midpoint temperature. That's not always gonna be the case. Okay, that's just a feature of this problem. Um Okay, so the temperature right here at the boundary 50.04. Now we know that temperature. That means that we can find a church, we can find of copper, we can find a church of aluminum. But remember that we said because we have no heat loss to the surroundings, that those are going to be equal. Okay, so we only need to calculate one of them, The other one is gonna be the same. Okay, so let's go ahead and do that calculation for aluminum. Okay, So h the heat current and the problem. Okay. And we'll call it H. A. Because we are looking at aluminum recall is watts per meter degrees Celsius times 0.86 m squared times the temperature T. Which we just found, 50. degrees Celsius minus zero degree Celsius, All divided by 0.4 m. Okay, this is the equation we had up here for the aluminum and we've just subbed in the value of T. We found and we get an H value of .5513. Okay. And for the unit we get watt per meter degree Celsius times meter squared times degrees Celsius case of the degrees Celsius, cancel. One of the meter terms cancel. And then we divide by meter again. So we're left with just the unit of what, which is what we want. Alright, So we're getting close, we found H. Okay. And let's go back and look at what we're trying to do. So we're trying to calculate the mass M. In order to do that. We need to find this value of Q. In order to find the value of Q. We need to find the value of H. We've done that. Okay, So now let's calculate the value of Q. All right. So recall that the heat transfer H. It's going to be equal to the quantity of heat Q divided by the time T. Which tells us that Q is equal to H times T. We found a church to be 220. watts. And the time is 10 minutes. Okay, now we want this in seconds. So we're gonna take 10 minutes. We're gonna multiply 60 seconds per minute. Mhm. And this is going to give us a q value of 100 and 32, 330.78. 30.78. And our unit is going to be watt second, which is alright, so we're close. We found the value of Q. And now recall that our mass M. That we're looking for is going to be equal to the quantity of heat Q divided by the latent heat of fusion LF. Okay? And in this case we're talking about water, we're melting water or sorry, we're melting ice into water. And so we're talking about the latent heat of fusion for water. So we get our value of Q 132330. jewels divided by 334. Okay. And this is a value you can look up in the table in your textbook or that your professor provided times 10 to the three. And the unit here is jewel per kilogram. The unit of jewels will cancel. We're left with the unit of kg, which is what we want for mass. We get 0. Kg. Okay. And so that is the amount of ice that will melt in minutes using that heat supplied by the Rod. Alright, So if we go back up to Our answer choices, we see that we have answer choice a 0. kg of ice will melt in 10 minutes using that heat supplied by the Rod. Thanks everyone for watching. I hope this video helped see you in the next one

Verified Solution

Key Concepts

Thermal Conductivity

Heat Transfer

Latent Heat of Fusion