Home
Class 11
PHYSICS
A solid sphere of copper weighs 1kg, Fin...

A solid sphere of copper weighs 1kg, Find the increase in its surface area when its temperature rises from `15^(@)C" to "500^(@)C`. Relative density of copper at `0^(@)C` is 8.39. `alpha=16.07xx10^(-6)//""^(0)C`

Text Solution

Verified by Experts

`r_(0) = [ (1000)/(4//3 pi) ]^(1//3) = 2.99 cm , r_(15) " at " 15^(0)C`
`r_(15) = r_(0) ( 1 + alpha Delta theta) = 2.99 (1 + 16.07 xx 10^(-6) xx 15) =2.99 1` cm
`therefore ` Area at `15^(@)C = A_(15) = 4 pi r_(15)^(2) = 112.4cm^(2)`
At `500^(0)` the radius is `r_(500)` = 2.99 `(1 + alpha xx 500)` = 3.014
`A_(500) = 4 pi r_(500)^(2) = 114.2cm^(2)`
Hence increase in surf ace area
`Delta A = A_(500) - A_(15) 114.2 - 112.4 = 1.8 cm^(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

When water is heated from 0^(@)C to 10^(@)C , its volume

A system absorbs 10kJ of heat at constant volume and its temperature rises from 27^(0)C " to " 37^(0)C . The DE of reaction is

The area of a circular copper coin increases by 0.4 % when its temperature is raised by 100^(@)C . The coefficient of linear expansion of the coin is :

Statement-1:- The temperature dependence of resistance is usually given as R=R_(0)(1+alpha Deltat) . The resistance of a wire changes from 100Omega to 150Omega when its temperature is increased from 27^(@)C to 227^(@)C . This implies that alpha=2.5xx10^(-3)//""^(@)C Statement- 2 :- R=R_(0)(1+alpha Deltat) is valid only when the change in the temperature DeltaT is small and DeltaR=(R-R_(0))lt lt R_(0)

The volume of a sphere is increasing at the rate of 1200 c.cm/sec. The rate of increase in its surface area when the radius is 10 cm is