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 . . . . . . ..

Options

A. 240
B. 360
C. 480
D. 216

Answer

D

Explanation

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`