Question

When the temperature of a copper coin is raised by 80?

Answer 1

1 answer · Physics 

 Best Answer

Hi, 
Let the initial diameter = d 
After heating to 80 C, diameter becomes = d + 0.2%d = d(1+0.2%) = d(1+0.002) = 1.002d cm 

a) area of the face of the coin increases by 0.6 percent 
Initial Area = pi*d^2 = pi*d^2 cm^2 
After Heating Area will be = pi(1.0002d)^2 = 1.004004pi*d^2 cm2 
% increase => change/original x 100 
% increase in area = (1.004004pi*d^2 - pi*d^2)*100/(pi*d^2) = (1.004004 -1)pi*d^2*100/(pi*d^2) = 0.004004*100*pi*d^2/(pi*d^2) = 0.4004 % 
so a) is not correct 

b)thickness increases by 0.6 percent 
As increase in diameter is 0.2 % which is linear expansion, and thickness will alse observe linear expansion, so it shall be 0.2% i.e., same as that of diameter and not 0.6 percent. 
So b) is also wrong. 

c)volume increases by 0.4 percent 
let initial thickness = t, and after heating thickness will be = 1.002 t 
Inital volume = pi*d^2*t cm^3 
Afetr heating volume will be = pi*(1.002d)^2*(1.002t) = (1.002)^3*pi*d^2*t = 1.006012008pi*d^2*t 
% increase in volume = (1.006012008pi*d^2*t - pi*d^2*t)*100/(pi*d^2*t) = (1.006012008 - 1)pi*d^2*t*100/(pi*d^2*t) = 0.006012008*100 = 0.6012008% 
so c) is also not correct 

d)coefficient of linear expansion of the material will be 0.25 x 10 to the power -4degree centigrade. 
Coefficient of linear expansion = change in length/change in temperature 
= (1.002 - 1)/80 = 0.002/80 = 0.25 x 10 to the power -4, 

Answer 2

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