Two solid copper spheres of radii `r_1=15cm` and `r_2=20cm` are both at a temperature of `60^@`C. If the temperature of surrounding is `50^@C`, then find
a. The ratio of the heat loss per second from their surfaces initially.
b. the ratio of rates of cooling initially.

Explanation

a. Ratio of heal loss `=(H_1)/(H_2)=((dQ//dt)_1)/((dQ//dt)_2)implies(H_1)/(H_2)=(KA_(1)(60-50))/(KA_(2)(60-50))` by using newton's law of cooling
`(H_1)/(H_2)=(A_1)/(A_2)=(r_1^2)/(r_2^2)=((15)/(20))^2=(9)/(16)`
b. The ratio of initial rater of cooling `=((d theta//dt)_(1))/((d theta//dt)_(2))`
we have `(H_1)/(H_2)=((r_1)/(r_2))^2implies(M_1s(d theta//dt)_1)/(M_2s(d theta//theta)_2)=((r_1)/(r_2))^2`
As the sphere have the same densities, the ratio of their masses is equal to the ratio of their volumes.
`implies((d theta//dt)_1)/((d theta//dt)_2)=((r_1)/(r_2))^2(M_2)/(M_1)=((r_1)/(r_2))^2((r_2)/(r_1))^2=(r_2)/(r_1)=(20)/(15)=(4)/(3)`.