The rate at which the metal cools in moving air is proportional to the difference of tempratures between the metal and air. If the air temperature is 290 ~K and the metal temperature drops from 370 ~K to 330 ~K in 10 minutes, then the time required to drop the temperature upto 295 ~K is

Step-by-step solution

(D) We know that dT dt =- k ( T - Tm ) , where k is proportionality constant, T = Temperature of body, Tm = Temperature of surrounding. dT dt =- k ( T -290) dT T -290 = - k dt | T -290|=- kt + C Initially, t =0, ~T =370 |370-290|=0+C C= |80| | T -290|=- kt + |80| When t =10, ~T =330 |330-290|=-10 k+ |80| [ [ 40 80 ] ]=-10 k k = 1 10 2 [|T-290|]=- 2 10 t+ |80| When T =295 , we write [295-290]]=- 2 10 t+ 80 ( 5- 80) 2 (-10)=1 ( 1 10 ) 2 (-10)=1 t= -4 2 2 (-10)=40