Question

A metal ball cools from `62^(@)C` to `50^(@)C` in `10min` and to `42^(@)C` the next ten minutes. What will be its temperature at the end of next ten minutes ?

Answer 1

Use `(theta_(1)-theta_(2))/(t)=K((theta_(1)-theta_(2))/(2)theta_(0))` ………`(1)`
to get the following equations
`(62-50)/(10)=K((62+50)/(2)-theta_(0))` and
`(50-42)/(10)=((50+42)/(2)-theta_(0))`……..`(2)`
Divide `2` by `1` and solve to get `theta_(0)`
`(8)/(12)=(46-theta_(0))/(56-theta_(0))`
`implies theta_(0)=26`.......`(3)`
Let after the next `10min` the temperture falls to then
`(42-theta)/(10)=K((42+theta)/(2)-26)`......`(4)`
From `(3)` using value of `theta_(0)=26` we get
`(12)/(10)-k(56-26)`
Divide `(4)` by `(5)` to get
`(42-theta)/(12)=(42+theta-52)/(2(56-26))`
`implies theta=36.7^(@)C`