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

Explanation

Rate of cooling, `(d theta)/dtprop` excess temperature over the surroundings.
In the first case `1/10prop40`,
In the second case `(d theta)/(dt)prop30`
divifing `(d theta)/(dt)//1/10=30/40`
`rArr(d theta)/(dt)=3/(4xx10)=3/40""^(@)C//sec`
Rate of heat radiation = `ms(d theta)/(dt)`
= Thermal capacity `xx(d theta)/dt=4600xx3/40=345J//sec.`