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

Explanation

Rate of cooling , `(d theta)/(dt) prop` excess temperature over the surroudings.
In the first case `(1)/(10) prop 40`, In the second case `(d theta)/(dt) prop 30` .
diving `(d theta)/(dt) //(1)/(10) =(30)/(4) " "therefore (d theta)/(dt) = (3)/(4 xx 10) =(3)/(40)""^(@)C// sec `
Rate of heat radition ` = ms ( d theta)/(dt) `= Thermal capacity `xx ( d theta)/(dt) = 4600 xx (3)/(40) = 345J//sec`