A typical car has 17 L of liquid coolant circulating at a temperature of 95 degrees celsius through the engine's cooling systemAssume that, in this normal condition, the coolant completely fills the 4.0 L volume of the aluminum radiator and the 13.0 L internal cavities within the steel engineWhen a car overheats, the radiator, engine, and coolant expand and a small reservoir connected to the radiator catches any resultant coolant overflowEstimate how much coolant overflows to the reservoir if the system is heated from 95 degrees celsius to 108 degrees celsiusModel the radiator and engine as hollow shells of aluminum and steel, respectivelyThe coefficient of volume expansion for coolant is beta410X10^-6 /celsius degrees.
Non stick side would be the dull sideMany aluminum foil products started to market with this information such as Reynolds WrapIn today's technology, the smoother and the shinier surface, the better for food substance to stick onThe dull surface such as your teflon, and other cookware consist of micro/nano particles coating that will reduce the surface area where food can stick, just like trying to stick a bandge to rough surface, it will easily peeledcoating can be anykinds of material, such as teflon, crystals, inert metals etc.
Use baking soda it reacts with it leaving a thin black layer over the foil making it look like iron.
Hiya! I have the same professor as you right now :) So we have two parts to this problem, one container with aluminum and another with steelWe need to find the delta(V) or change in volume for both parts using: dV (V_o)(beta)(delta_T) beta(aluminum) 75 x 10^-6 beta(steel) 35 x 10^-6 Aluminum: dV (4.0L)(75 x 10^-6)(108 - 95) .0039L Steel: dV (13.0L)(35 x 10^-6)(108-95) .005915L Now the NEW volume of each respective container is 4.0039L and 13.005915L, giving a new total volume of 17.009815LWe'll come back to this number(You could have done this next step first but whatever) Now we look at the volume expansion of the coolant: dV (17.0L)(410 x 10^-6)(108-95) .09061L What we want to know is how much spills over to the reservoir, which can be found by comparing the two new volumes we just calculated: The coolant is now at a new volume of 17.09061L and the carrying capacity of the containers total 17.009815LThere is more coolant than space available! so just subtract it: 17.09061L - 17.009815L .080795L, or 80.795mL, or rounded to two sigfigs 81mL :)
