Question

If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of the hole when the rivet is cooled to - 78.0 ∘C, the temperature of dry ice? Assume that the expansion coefficient of aluminum is 2.4×10−5(∘C)−1 and remains constant.

Answer 1

Answer:

Explanation:

diameter at 23 degree celcius is 4.511 mm

GIVEN DATA:

diameter of hole  = 4.500 mm

T_1 = 23.0 Degree celcius

T_2 = - 78.0 Degree celcius

expansion coeffcient of Al = 2.4*10^{-5} (degree celcius)^{-1}

diameter at 23 degree celcius is given as

$d = d_o (1+ \alpha \delta T)$

 $= 4.5 (1+2.4*10^{-5} *(23-(-78)))$

   = 4.511 mm

diameter of rivet after change in temperature is given as

$d= d_o + \Delta d$

 $= d_o(1+ \alpha \delta T)$

$= 0.4500 *(1+2.4*10^{-5} *(23-(-78)))$

= 0.4511 cm