00:01
So here in this question, first we need to find the surface area of copper rod.
00:04
So the surface area, surface area of copper rod, of copper rod is given by a and that is equals to 2 pi r h which is the surface area of cylinder ignoring the ends.
00:23
Now, here r is the radius and h is the height of the cylinder.
00:27
In this case, the diameter is 2 centimeters.
00:30
So radius is diameter by 2 and that is equals to 2 centimeter by 2 that is 1 centimeter.
00:38
Now let's assume the length of the rod is l meters then the surface area is a is equals to 2 pi into 0 .01 meters into l.
00:49
Next we need to find the heat transfer rate the heat transfer rate is given by heat transfer rate heat transfer rate is given by q and that is equals to h a delta t where h is the heat transfer coefficient a is the surface area and delta t is the temperature difference between the rod and the air stream now plugging in the values q is equals to h that is 200 watts per meter square kelvin into 2 pi into 0 .01 meters into l into temperature difference that is 100 degree celsius minus 20 degree celsius that is 80 degree celsius.
01:35
Now we need to find the amount of heat that needs to be removed from the copper rod to cool it down to average temperature of 25 degrees celsius.
01:45
Now the heat capacity is 300.
01:49
Heat capacity is 385 joules per kg per kelvin...