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Mechanical Properties Of Fluids

Question
CBSEENPH11019845

A solid sphere of radius R, made from a material of density ρ and specific heat s is heated to temperature T and placed in an enclosure at temperature T0 Show that the:

(i) rate of loss of heat varies directly with square of radius and independent of specific heat and density of the material,

(ii) rate of cooling varies inversely with radius, specific heat and density.

Solution

i) According to Stefan’s law, the rate of loss of heat by a body is, 
negative space space space space dQ over dt equals space eAσ space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space
space space space space space space dQ over dt space equals space minus straight e 4 πR squared straight sigma space space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space
rightwards double arrow space space dQ over dt equals space minus space left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis space straight R squared
From the above relation, it is clear that the rate of loss of heat varies directly with the square of radius and independent of specific heat of material and density of material. 
ii) Now, the rate of cooling varies inversely with radius, specific heat, and density. 
space space space space space space space dQ over dt space equals space msdT over dt

rightwards double arrow space space dT over dt space equals space minus left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket straight R squared over ms

space space space space space space space space space space space space space space space equals space minus space left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket space fraction numerator 3 straight R squared over denominator 4 πR cubed ρs end fraction

space space space space space space space space space space space space space space equals space minus space left square bracket 3 eσ space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket 1 over Rρs

rightwards double arrow space dT over dt space proportional to space 1 over Rρs
Hence, the result.