A body cools from ` 62 ^@ ` C to ` 54 ^@ ` C in 10 minutes and to ` 48^@ ` C in the next 10 minutes. Find the temperature of the surroundings.

Explanation

` theta _ 1 = ( 62 + 54) /( 2 ) = 58 ^@ ` C
` theta _ 2 = ( 54 + 48 ) /( 2 ) = 51 ^@ ` C
` ( d theta _ 1)/( d t ) = ( 62 - 54 ) /( 10 ) = 0.8 ^@` C/min
` ( d theta _ 2 ) /( dt ) = ( 54 - 48 ) /( 10 ) = 0.6 ^@ ` C/min
According to Newton's law of cooling :
` ( d theta _ 1 ) /( dt ) = k ( theta _ 1- theta _ 0 ) `
and ` " " ( d theta _ 2) /( dt ) =k ( theta _ 2- theta _ 0 ) `
` therefore ( ( d theta _ 1 ) /( dt ) ) /(( d theta _ 2 ) /( dt )) = ( k ( theta _ 1 - theta _ 0 ) ) /( k ( theta _ 2 - theta _ 0 )) = ( theta _ 1 - theta _ 0 ) /( theta _ 2 - theta _ 0 ) `
` therefore " " ( 0.8) /( 0.6 ) = ( 58 - theta _ 0 ) /( 51 - theta _ 0 ) `
` therefore 0.8 ( 51 - theta _ 0 ) = 0.6 ( 58 - theta _ 0 ) `
` 40.8- 0.8 theta _ 0 = 34.8 - 0.6 theta _ 0 `
` 0. 8 theta _ 0 - 0.6 theta _ 0 = 40.8 - 34.8 `
` 0.2 theta _ 0 = 6 `
` theta _ 0 = ( 6 ) /( 0.2 ) = 30 ^@ ` C
` therefore ` Room temperature ` theta _ 0 = 30 ^@ ` C