The rate of cooling of water in a calorimeter is one-tenth of a degree per second when its teniperature is `40^(0)` C above that of the sorroun- dings. What is the rate of cooling and lteat per second when the excess temperature over the surroudings is `30^(@)`C ? Thermal capacity of water and calorimater is 4600 J`//^(0)` C .

Explanation

`(d theta)/(dt ) prop` excess temperature over the sorroundings .
in the first case `(1)/(10) prop 40`,
In the second case `(d theta)/(dt) prop` 30
dividing `(d theta)/(dt) / (1)/(10) = (30)/(40)`
`rArr (d theta)/(dt) = (3)/(4 xx 10) = (3)/(40) " "^(@) C ` / sec
Rate of heat radiation = ms `(d theta)/(dt)`
= thermal capacity `xx (d theta)/(dt)`
`= 4600 xx (3)/(40) = 3451` J / sec.