Newton\'s law of cooling states that for small temperature differences, if a bod
ID: 2010061 • Letter: N
Question
Newton's law of cooling states that for small temperature differences, if a body at a temperature T1 is in surroundings at a temperature T2. the body cools at a rate given by Delta Q/ Delta T = K(T1 - T2) where K is a constant. It includes the effects of conduction, convection, and radiation. That this linear relationship should hold is obvious if only conduction is considered. Show that it is also approximately true for radiation by showing that equation Delta Q/ Delta T = e sigma A(T4 1 - T4 2) reduces to Delta Q/ Delta T = e sigma A(T4 1 - T4 2) = constant Times (T1 - T2) (Hint: use the formula for the difference of squares).Explanation / Answer
Q / T = e A ( T1^4 - T2 ^4 ) ;
Q / T = e A [ ( T1^2 )^2 - ( T2 ^2)^2 ] ;
Q / T = e A [ T1^2 + T2^2 ] [ T1 ^2 - T2 ^2 ] ;
Q / T = e A [ T1^2 + T2^2 ] [ T1 + T2 ] [ T1 - T2 ] ;
since there is very small difference between T1 and T2 instantaneously ,
T1 + T2 = 2T2 , and [ T1^2 + T2^2 ] = 2 T2^2 , so
Q / T = e A [ 2 T2^2 ] [ 2 T2 ] [ T1 - T2 ] ;
Q / T = 4 e A * T2^3 * [ T1 - T2 ] ;
Q / T = [ 4 e A * T2^3 ] * [ T1 - T2 ] ;
Q / T = k * [ T1 - T2 ] ; for small difference between T1 and T2 insrtantaneously
hence proved