Common outdoor thermometers are filled with red-colored ethyl alcohol. One thermometer has a 0.40-mm-diameter capillary tube attached to a 9.0-mm-diameter spherical bulb. On a 0°C morning, the column of alcohol stands 30 mm above the bulb. What is the temperature in °C when the column of alcohol stands 130 mm above the bulb? The expansion of the glass is much less than that of the alcohol and can be ignored.

Video transcript

Hey everyone. So this problem has to do with volumetric thermal expansion. Let's see what it's asking us. A science use a scientist uses a thermometer containing isopropyl alcohol with a volumetric thermal expansion coefficient denoted by beta of 7.9 times 10 to the negative four per Kelvin to monitor a temperature sensor experiment. The thermometer has a 0.5 millimeter diameter capillary tube and an eight millimeter diameter spherical cap or spherical bulb at zero degrees Celsius. The alcohol column is 44 millimeters above the bulb, determine the temperature when the column reaches 145 millimeters. Considering the ga the glass expansion is negligible compared to the alcohol. So our multiple choice answers here are a 48 degrees Celsius B 95 degrees Celsius, C 72 degrees Celsius or D 88 degrees Celsius. So the first thing I wanna do here to solve the problem is draw a little diagram to make sure we understand the um problem that is or what the problem is asking us. So we have this thermometer where the spherical bulb at the bottom has a diameter of eight millimeters and then the diameter of this tube is 0.5 millimeters. And so the alcohol initially is at this point where it is 40 millimeters above the bulb, and then eventually it will continue to rise to a total point of 145 millimeters. And so that height difference is caused by the thermal expansion of the alcohol. And so that is how we are going to solve for the temperature. So the equation that we can use for the change in volume as a uh function of the change in temperature is going to be delta V is equal to V I multiplied by beta, that thermal expansion coefficient multiplied by delta T. So another way to write that if you want to expand out these deltas is VF minus V I is equal to V I multiplied by beta multiplied by TF minus T I. And so TF at final temperature is what we are going to be solving for in the problem. And so if we isolate that on the left hand side of the equation, we get TF is equal to VF minus V I divided by V I multiplied by beta plus T I. And so then from there, we need to look at each of these terms, whether we have them and whether we can solve for them. So the problem gives us T I of zero degrees Celsius, it gives us the temperature thermal co temper sorry thermal thermal expansion coefficient. Excuse me of 7.9 times 10 to the negative four per Kelvin or Kelvin to the negative one. And so then we, we what we don't have are our volumes V I and VF but we are given um dimensions that we can use to solve for those volumes. And so our initial volume is going to be the volume of the bulb plus the volume of the tube when the height is only at 40 millimeters. And so the volume of the bulb, it's a sphere. We can recall that that is four thirds pi are squared where the radius is half of the diameter. So that will be four times 10 to the negative 3 m cubed. So that um the volume of a sphere, four thirds pi R cubed. And then we just plugged in that R which we got from our diameter plus the volume of a of the tube. So we can recall the volume of a cylinder is pi R squared H. So we'll have pie again, our radius is going to be half of that diameter. So point to five times 10 to the negative 3 m and that quantity is squared and then multiplied by H. So the initial um height is millimeters or 40 times to the negative 3 m me rewrite that. So it's all stays on the page 10 to the negative 3 m. And so we plug that in to our calculator and we get an initial volume of 2.7 59 times 10 to the negative seven cubic meters. And now we're going to do the same thing for our final volume where the volume of the sphere isn't changing. And so the only thing that's changing here is the, well, it is the height of that uh of the in the volume of the tube part of that equation. So that will look like pi multiplied by 0. times 10 to the negative 3 m that quantity squared that stays the same. And then the only thing that changes here is that height term. So now we have 145 times 10 to the negative m. So we plug that in and we come up with 2.966 times 10 to the negative 7 m cubed. And so now when we look at our initial equation for final temperature, we do have all of the values we need and it becomes one last final plug and chug step to get to that number. So we have VF 2.966 times 10 to the negative 7 m cubed minus V I 2.759 times 10 to the negative 7 m cubed all divided by V I 2.759 times 10 to the negative 7 m cubed multiplied by our thermal expansion coefficient beta 7. times 10 to the negative four degrees Kelvin. Now the question asked or the answers are in degrees Celsius and our initial temperature is also in degrees Celsius. So we can recall here that as a unit, a degree Celsius is equal to a degree Kelvin. And so we can just simplify that by rewriting that as per degree Celsius. And so we'll add that to our t initial or zero degrees Celsius and solve. And when we plug all of that in, we get 95 degrees Celsius. So that is the final answer for this problem. And it aligns with answer choice B that's all we have for this one. We'll see you in the next video.