A hot metal sphere cools from `60^(@)C " to " 52^(@)C` in 5 minutes and from ` 52^(@)C " to"44^(@)` C in the next 7.5 mintues. Determine its temperture after the next 10 minutes.

Explanation

Let `theta_(0)` be the room temperature ann `theta` be the final temperature of the metal sphere.
During the first 5 min. The rate of cooling is ` (60-52)/5 ^(@)C` min and the average temperature excess is `((60 +52)/2 - theta_(0))` . According to Newton's law of cooling .
Rate of cooling ) ` prop` (average temperature excess)
` (60-52)/5 prop ((60 +52)/2 - theta_(o)) `
`(60-52)/5 = k ((60+52)/2 - theta_(0))`
` 8/5 = k (56-theta_(0))`
Similarly , during the next 7.5 minutes,.
` (52-44)/7.5 = k ((52+44)/2- theta_(0))`
` 8/7.5 = k (48- theta_(0))`
Dividing Eq. (1) by Eq (2)
` (56- theta_(0))/(48-theta_(0)) = 7.5/5 = 3/2`
` 112 - 2 theta _(0) = 144 - 3 theta_(0) " " therefore theta_(0) = 32^(@) C`
Subsituting for ` theta_(0) ` in Eq. (1) `8/5 = k (56-32)`
` k = 8/(5 xx 24) = 1/15 min^(-1)`
During the final 10 minutes the rate of cooling is `((44-theta)/10) ` and the average temperature excess is `((44 +theta)/2 - theta_(0)) `
According to Newton's law of cooling
` (44-theta)/10 = k ((44 + theta)/2 - theta_(0))`
Substituting ` theta_(0) = 32 and k = 1/15` we get
` (44-theta)/10 = 1/15 ((44 + theta)/2 -32)`
Simplifying 132 - ` 3theta = 44 + theta - 64`
` theta = 38^(@) C`