Question

a sphere and a cube both made of copper have equal volumes and are blackened these are heated to same temperature and are allowed to cool under same surrounding the ratio of their rates of heat loss is

Answer 1

The ratio of there heat loss of sphere to cube is 0.80, which means cube cooled faster than sphere.

The heat loss due to radiation is given by Stefan's law

Q=σA(ΔT^{4})

Therefore, Under similar condition, the heat loss is directly proportional to the surface area alone.

Volume of cube=Volume of sphere

Let, the side of the cube is a, and radius of the sphere is r, than

a^{3}=(4/3)πr^{3}

a=1.611r

Now the heat loss ratio of sphere to volume is calculated as

$\frac{Q_sphere}{Q_cube}$=$\frac{4πr²}{6a²}$

=$\frac{4πr²}{6*(1.611r)²}$

$\frac{Q_sphere}{Q_cube}$=0.80