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.

The rate of cooling of water in a calorimeter is one-tenth of a degree

1.	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.
Answer» Solution :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.`