Question

Your answer is partially correct. Try again. Consult Conceptual Example 5 for background pertinent to this problem. A lead sphere has a diameter that is 0.050% larger than the inner diameter of a steel ring when each has a temperature of 26.8 °C. Thus, the ring will not slip over the sphere. At what common temperature will the ring just slip over the sphere? Steel ring Lead ball Number 155.8 umiestne The number or significant digits is set to 3; the tolerance is +-1 in the 3rd significant dit

Answer 1

The change in the diameter of the lead sphere is given by Δddo Q ΔΤ where a of lead 29 x 10-6 (oC)-1 is the linear thermal expansion coefficient The diameter of the lead sphere after the change (original diameter plus change in diameter) is given by dolAd dodoioiAT dol (1aAT) Similarly the inner diameter of the steel ring after change is given by dos Adsdos(1+ asAT) where as12 x 10- (°C)-1 is the linear thermal expansion coefficient of steel When the ring just slips over the sphere, the inner diameter of the ring must be equal to the diameter of the sphere. That is dolAd dos Adg (or) do1 (1aAT) dos(1asAT) This equation can be rewritten for the change in temperature AT as dos - dol AT = andol agdos But the diameter of the lead sphere is 0.050% greater than the inner diameter of steel ring. That is 0.050 dot dos x dos dos(10.0005)= 1.0005 x dos 100

T_f = 29.4 + 26.8 = 56.2°C

So the change in temperature can be written as dos(1 1.0005) dos (1.0005 x a dos 1.0005 x dos ΔΤ. 1.0005 x ados - asdos -0.0005 (1.0005 x a as) as) Substituting the values of linear thermal expansion coefficients -0.0005 -0.0005 ΔΤ. (1.0005 x 29 x 106 29.4° C (or) 12) x 10-6 17.0145 AT Tf T =-29.4° C