A hot liquid in beaker at 80^(@)C cools down to 70^(@)C in 2 minutes in an atmosphere of 30^(@)C temperature, then time taken to cool it down from 60^(@)C to 50^(@)C is . . . . . . ..

A hot liquid in beaker at 80^(@)C cools down to 70^(@)C in 2 minutes i

1.	A hot liquid in beaker at 80^(@)C cools down to 70^(@)C in 2 minutes in an atmosphere of 30^(@)C temperature, then time taken to cool it down from 60^(@)C to 50^(@)C is . . . . . . ..
Answer» 240 360 480 216 Solution :TEMPERATURE of surroundings `T_(s)=30^(@)C` Average temperature in SITUATION - 1 `T_(1)=(80+70)/(2)=75^(@)C` Average temperature in situation - 2 `T_(2)=(60+50)/(2)=55^(@)C` According to Newton.s COOLING rate, `(dT)/(dt)=K.(T-T_(s))` `:.(dT)/(K.)=(T-T_(s))dt` But `(dT)/(K.)` is constant `:.(T-T_(S))dt="constant"` `:.(T_(1)-T_(s))t_(1)=(T_(2)-T_(s))t_(2)` `:.t_(2)=((T_(1)-T_(s))/(T_(2)-T_(s)))t_(1)` `=((75-30)/(55-30))2xx60` `=(45)/(25)xx120` `:.t_(2)=216s`