A cup of coffee cools from `150^(@)F` to `144^(@)` F m 1 min in a room at `72^(@)F.` How much time will the coffee take to cool from `110^(@)F` to `104^(@)F` in the same room ?

Options

A. `1.55` min
B. `2.14` min
C. `2.89` min
D. `3.35` min

Answer

B

Explanation

Given,
A cup of coffee cools,
`T = 150^(@)F`
`T _(2) = 144 ^(@) F`
`t _(1) =1m`
Room temperature, `T_(0) = 72^(@) F`
`T_(2) = 110^(@) F`
` T_(4) = 104^(@) F`
According to Newton's law of cooling.
For ` T _(1) , T_(2) and t _(1)`
`(mc (T_(1) - T_(2)))/(t _(1)) = K ((T_(1) + T_(2))/(2)-T_(0)))" "...(1)`
`(mc (150 - 144))/(1) =K ((150 + 144)/(2) - 72)`
`6 cm C = K (147 -72)`
` 6mC = 75 k`
For `T_(3), T_(4) and t _(2)`
`(mc (110 - 104)/(t _(2))) =K ((110 + 104)/(2)- 72)`
`(6 mC)/(t _(2)) = 35K`
`t _(2) = ( 6mC)/(35 K ) " "...(2)`
Subsituating equation (2) in equatin (3)
`t _(2) = (75K)/(35k)= 2.14min`
`therefore` Time taken by coffee to cool `= 2.14 min`