A solid metallic sphere of diameter 20 cm and mass 10 kg is heated to a temperature of `327^@C` and suspended in a box in which a constant temperature of `27^@C` is maintained. Find the rate at which the temperature of the Sphere will fall with time. Stefan's constant`=5.67xx10^-8W//m^(2)//K^(4)` and specific heat of metal `=420J//kg//^(@)C`.

Explanation

The rate of loss heat by the sphere is given by
`(dQ)/(dt)=sigmaA(T^4-T_0^4)`
Where A is the surface area of the sphere`=4pir^2`, with `r=10cm=0.1m`,`T=327^@C=600K` and `T_0=27^@C=300K`
Thus `(dQ)/(dt)=5.67xx10^-8xx4pixx(0.1)^2xx{(600)^4-(300)^4}=866J//s`
Now `dQ=msdT`, where dT is the fall in temperature in time dt.
`(dQ)/(dt)=ms(dT)/(dt)`
or `866=10xx420xx(dT)/(dt)`
or `(dT)/(dt)=(866)/(4200)=0.206^@C//s`