A body initially at `80^(@)C` cools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What is the temperature of the surroundings?

Options

A. `24^(@)`
B. ` 28^(@)`
C. `22^(@)`
D. `25^(@)`

Answer

A

Explanation

According to newton's law of cooling
` (80-54)/5 = alphi [ (80 +64)/2]` -T`
` and (80 -52)/10 = alpha [(80 +52)/2] -T`
Where T is the temperature of the surrounding solving Eq. (i) and (ii) we get