4. a) The inner diameter of a steel ring is 2.0000 cm, and the diameter of an aluminumdisk is 2.0100 cm. Both are at 430C. At what common temperature will the disk fitprecisely into the hole in the steel ring?b) If after the aluminum disk is fitted precisely into the hole the two metals are thetemperature is changed to 200 C, what is the stress in the steel ring?
Depends on what you're using it for. For cooking, we've recently discovered old fashioned cast iron . . . and love it! If the choice is between steel and aluminum, however, I'd use stainless steel. But since your question is under home and garden, maybe you just want to carry something in a pan, in which case aluminum will accomplish the same thing and be a lot lighter.
Steel ring: - Inner diameter: 2.00 cm at 430-deg C - thermal expansion coefficient: 17.3e-6 = s Aluminum ring: - Outer diameter: 2.01 cm at 430-deg C - thermal expansion coefficient: 23.1e-6 = a a) The aluminum ring is too large. If they are both cooled, the aluminum ring will shrink at a faster rate, so at some cooler temperature, there will be a match. What we need: R_new = 2.01*(1+ a*dT) = 2.00*(1 + s*dT) 2.01 + (2.01*a)dT = 2.00 + (2.00*s)dT 0.01 = dT*(2*s - 2.01*a) dT = 0.01*e6/(2*17.3 - 2.01*23.1) = -8.5e2 (deg-C) = -850 (deg-C) Hmm, so we need a reduction in temperature by 850 deg. But we started at T-celsius = 430 = T-kelvin = 803. A reduction by 850 deg gives us -47 deg-K. But that's impossible! Either the coefficients of thermal expansion are extremely different at 430-C than at 20-C (quite possible) or this degree of accommodation is not possible. b) This cannot be answered, based on the impasse in a).
