A pan filled with hot food cools form `94^(@)C` to `86^(@)C` in 2 minutes . When the roo temperature is `20^(@)C`. How long will it cool from `74^(@)C` to `66^(@)C` ?

Options

A. 2 minutes
B. 2.8minutes
C. 2.5 minutes
D. 1.8 minutes

Answer

d

Explanation

According to Newtons.s law of coolinig
`(d theta)/(dt) = (theta_(1)+theta_(2))/(2)- theta_(0)` ………(i)
Here `(d theta)/(dt)= (94-86)/(2)=4^(2)Cmin^(-1)`
`theta_(1)=94^(2)C, theta_(2)=86^(@)C,theta_(0)=20^(@)C`
Using eqn (i)
`4=K(94+86)/(2)-20=70K`
`K= (4)/(70)`
For fall in temperature from `74^(@)C` to `66^(@)C`
Using eqn (i)
`(74-66)/(t) = (4)/(70)(74+66)/(2)-20= (4)/(70)xx50`
`t= (8xx7)/(4xx5)=2.8`mintutes