A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to ±1 mm) at 20°C. If she measures the distance between two stakes to be 65.175 m on a 3°C day, does she need to add or subtract a correction factor to get the true distance? How large, in mm, is the correction factor?

Video transcript

Hey, everyone. So this problem is dealing with linear thermal expansion. Let's see what it's asking us. We have a geologist studying stalagmite sediment rays in a 10 degrees Celsius cave that uses a 300 centimeter long brass measurement tool calibrated at 25 degrees Celsius. The geologist quantifies the stalagmite sediment to be 37.55 centimeters long. How should the geologist correct that value before publishing the results were asked to look at the change in this measurement tool due to the temperature. So our multiple choice answers here are a to subtract 110 micrometers B subtract 0.51 centimeters D sorry C add 110 micrometers or D add 0.51 centimeters. And so the key to solving this problem is we're calling our equation for the change in the length due to thermal expansion which is delta L is equal to alpha L delta T where alpha is our coefficient of linear thermal expansion that is a constant based on the material. In this case, it is brass and we can recall that that coefficient of linear thermal expansion of brass is 1. times 10 to the negative five per degree Celsius. Our length, our initial length is given as 37.55 centimeters. So, keeping everything in standard units, that's going to be 0.3755 m. And then our delta T is equal to our final temperature minus our initial temperature. And so the final temperature is the temperature at which the measurement was taken. So that's 10 degrees Celsius. Our initial temperature is the temperature at which the tool is calibrated 25 degrees Celsius. And therefore our delta T is negative degrees Celsius. And so from there, it is a plug and chug to find a change in length caused by this change in temperature. So we have 1.9 times 10 to the minus five per degree Celsius multiplied by 0.3755 m multiplied by negative 15 degrees Celsius. And so when we plug that into our calculator, we get a change in length or delta L of negative 1.1 times 10 and negative 4 m which is equal to negative 110 micrometers. So because it is a negative value, it means we're going to subtract that length of 110 micrometers and that aligned with answer choice. A so A is the correct answer for this problem. That's all we have for this one. We'll see you in the next video