A bdoy initially at `80^(@)C` ools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What will be the temperature (in `""^(@)C`) after 15 minutes?

Explanation

If `T_(0)` be the temperature of the surroundings, then
`-(dT)/(dt)=k(T-T_(0))`
or `(80-64)/5=k((80+64)/2-T_(0))`…………I
and `(80-52)/10=k((80+52)/2-T_(0))`………ii
After simplifying above equations we get `T_(0)=16^(@)C`
If T. be the temperature after 15 minutes, then
`(80-T.)/15=k((80+T.)/2-16)`...........ii
After simplify above equations we get `T_(0)=16^(@)C`
If T. be the temperature after 15 minutes then
`(80-T.)/15=k((80+T.)/2-16)`..........iii
From equation i and iii we get
`T.=43^(@)C`