You are boiling pasta and absentmindedly grab a copper stirring spoon rather than your wooden spoon. The copper spoon has a 20 mm ×1.5 mm rectangular cross section, and the distance from the boiling water to your 35°C hand is 18 cm. How long does it take the spoon to transfer 25 J of energy to your hand?

Video transcript

Hey, everyone. This problem is dealing with conductive heat transfer. Let's see what it's asking us. We are planning to roast meat on a fire. The iron rod that will be used to hold the meat is to be placed horizontally above the fire. But one end of the rod slipped off of the ring and touched the fire where the with the other end still touching the ring. So this situation is shown in the figure below the rod in the slanting position where the radius of the rod is given as 40 millimeters and the length is 55 centimeters. The temperature of the fire at the one end of the rod is 100 degrees Celsius. And we're asked to determine the time required to transfer 160 joules of energy from the fire to the ring that the other end of the rod is touching. We are told to assume the surrounding temperature to be 30 degrees Celsius, which means the temperature of the rod is 30 degrees Celsius. So our multiple choice answers here in units of seconds are a 3.1 B 4.2 C 5.3 or D 6.2. So this is a pretty straightforward problem as long as we can recall our conductive heat transfer equation. And we're looking at the rate of heat transfer. So that will be Q divided by delta T equals K, our thermal conductivity constant multiplied by A, the area multiplied by delta capital T, which is our change in temperature all divided by D or the distance that the um energy is being conducted across. And so solving for delta lower case T, which is our time, we have an equation of Q D divided by K A delta uppercase T. And let's take each of those terms, one by one Q was given to us in the problem as Jews D was given as centimeters. We want to keep everything in standard units. So I'm going to rewrite that as 550.55 m K or thermal conductivity constant for iron is 79.5 watts per meter Calvin. In our area of a circle, we can recall is pi R squared. And so the radius was given as millimeters, sorry, 40 millimeters. And so that becomes pi multiplied by 0.4 m. Again, keeping in standard units squared which equals 5.3 times 10 to the negative third meters squared. So that's our surface area. And then delta T delta upper case T for our change in temperature, we have the heat of the fire degrees C minus the heat of the um ring at 30 degrees C and so that equals 70 degrees C. So now all we have to do is plug all of those values into our delta T equation. So delta T equals Q Jews multiplied by 1600.55 m divided by what? Sorry, 79.5 watts per meter. Kelvin multiplied by 5.3 times 10 to the negative third meters squared multiplied by 70 degrees Celsius. Now, we're just looking at units, we don't need to worry about the difference between Kelvin and Celsius because one unit measure of a degree Kelvin is equal to one unit measure of a degree Celsius. So those units will cancel here. We plug that into our equation and we get a time of 3.1 seconds. So that is the correct answer for this problem. And that aligns with answer choice A so that's all we have for this one. We'll see you in the next video.