A body cools from `60 ^(@)C` to` 40^(@)C` in 7 minutes . What will be its temperature after next 7 minutes if the temperature of its surrounding is `10^(@)C`?

Explanation

In first case:
Initial temperature , `0_(1) =60^(@)C`
Final temperature , `0_(2) =40^(@)C`
Time of cooling , `l_(1) =7 `minutes `7xx60=420s`
Temperature of surroundings , `0_(0)=10^(@)C`
From Neton`s low of cooling, we can write, `(d0)/(dt)=K[(0_(1)+0_(2))/(2)-0_(0)]`
`rArr (60-40)/(420)=K [(60+40)/(5) -10]`
`rArr (20)/(420)=Kxx40 ...............(1)`
In second case :
Initial temperature, `0_(1) =40^(@)C`
Time of cooling , `t_(2)=7` minutes`=420s`
Again, from Newton`s lawof cooling we can write, `(40-0)/(420)=K [(40+0)/(2)-10].............(2)`
on solving equations (1) & (2) we get `0=28^(@)C`