let r is the radius of the pot.
given
V = 0.75 L
= 0.75*10^-3 m^3
use, V = (4/3)*pi*r^3
0.75*10^-3 = (4/3)*pi*r^3
==> r = 0.05636 m
surface area, A = 4*pi*r^2
= 4*pi*0.05636^2
= 0.03992 m^2
T1 = 100 C = 100 + 273 = 373 K
T2 = 20 C = 20 + 273 = 293 K
e = 0.6
using stefan-Boltzman's law
The rate of heat loss by radiation,
H = sigma*A*e*(T1^4 - T2^4)
= 5.67*10^-8*0.03992*0.6*(373^4 - 293^4)
= 16.3 W <<<<<<<<------------------------Answer
A spherical pot of hot coffee contains 0.75 L of liquid (essentially water) at an initial...
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