Question
A brass plug is to be placed in a ring made of iron. At 20 degrees Celsius, the diameter of the plug is 8.737 cm and that of the inside of the ring is 8.723 cm.
They must both be brought to what common temperature in order to fit? (ANSWER: -210 degrees Celsius)
What if the plug were iron and the ring brass?
(I got -208 degrees Celsius, but that's incorrect)
They must both be brought to what common temperature in order to fit? (ANSWER: -210 degrees Celsius)
What if the plug were iron and the ring brass?
(I got -208 degrees Celsius, but that's incorrect)
Answers
(brass diameter)*(1+(brass coefficient)*(change in brass temperature))
=
(iron diameter)*(1+(iron coefficient)*(change in iron temperature))
where:
brass coefficient = 19*10^-6 K^-1
iron coefficient = 12*10^-6 K^-1
=
(iron diameter)*(1+(iron coefficient)*(change in iron temperature))
where:
brass coefficient = 19*10^-6 K^-1
iron coefficient = 12*10^-6 K^-1
Nevermind. The answer is 250 degrees Celsius. Duh!
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