how to find the rate of cooling of a sphere of twice radius?

Question

It is given that a solid copper sphere cools at the rate of 2.8 degree C per min when the temperature is 127 degree C. Now how to find the rate of cooling of another copper sphere of twice radius at 327 degree in room of 27 degree Celsius?

Answer 1

Equate the Stefan–Boltzmann law pedagogically to

P = dQ/dt =c*m*dT/dt

= σ(T^4-To^4)*4*pi*r^2 = c*(4/3)*pi*r^3*rho

Work the math, a bunch of junk cancels, and I end up with:

[((600^4-300^4)/(2/3)) /

((400^4-300^4)/(1/3))]*2.8

= 9.72 °/min

Answer 2

A bit more complicated than that.

Cooling rate is a function of 3 things: Area, dT, and volume

The first & 3rd are functions of R² & R³ respectively

Thus, letting Θ = rate of heat loss,



Θ2 = Θ1*[R2/R1]²*[R1/R2]³*dT2/dT1 = Θ1*[R1/R2]*dT2/dT1

Θ2 = 2.8*[1/2]*(327 - 27)/(127 - 27) = 4.2C°/min

Answer 3

Rate of cooling is proportional to the difference in temperature and the surface area.



An object with a doubled radius would have a quadrupled surface area.



R = k * dT * A

k = R / (dT * A)

k = (2.8C/min) / (100C * 1)



R = k * dT * A

R = ((2.8C/min) / (100C)) * (300C * 4)

R = (2.8C/min) * 12

R = 33.6C/min

Answer 4

Hey thanks for your answer. But the answer must be from these options : a. 5.6 degree C/min b. 7.8 degree C/min c. 9.7 degree C/min d. 11.4 degree C/min .and also it would be better if you could explain the relation of rate of cooling with area. Thanks.