A solid body X of heat capacity C is kept in an atmosphere whose temperature is `T_(A)=300K` At time t = 0, the temperature of X is `T_(0)=400K`. It cools according to Newton’s law of cooling. At time `t_(1)` its temperature is found to be 350 K.
At this time `(t_(1))` the body X is connected to a large body Y at atmospheric temperature `T_(A)`through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time `t=3t_(1)` (in Kelvin)___________.

Answer

`[300+ 12.5 e^((2KAt_(1))/(CL))]K`

Explanation

In the first part of the equation `(tle t_(1))`
At `t=0 , T_(X) = T_(0) = 400K"and at" t=t_(1) T_(X) = T_(1) = 350K`
Temperature of atmosphere , `T_(A) = 300K` (constant)
This cools down according to Newton's law of cooling.
Therefore, rate of cooling `alpha` temperature difference.

`therefore (-(dT)/(dt)) = k(T-T_(A)) implies (dT)/(T-T_(A)) = -k.dt`
`implies underset(T_(0))overset(T_(1))int(dT)/(T-T_(A)) = -kunderset(0)overset(t_(1))intdt implies ln((T_(1) - T_(A))/(T_(0) - T_(A)))= -kt_(1)`
`implies kt_(1)= -ln((350-300)/(400-300))implies kt_(1) = ln(2)`
In the `II^(nd)` part , body `X` cools by radiation (according to Newton's law) as well as by conduction `(t gt t_(1))`.

Therefore , rate of cooling = (cooling by radiation) + (cooling by conduction)
In conduction `(dQ)/(dt) = (KA(T-T_(A))/(L) = C(-(dT)/(dt))`
`therefore (-(dT)/(dt)) = (KA)/(LC)(T-T_(A))`
where `C` = heat capacity of body `X`
`therefore (-(dT)/(dt)) = k(T-T_(A)) + (KA)/(CL) (T-T_(A)) ..... (ii)`
`(-(dT)/(dt)) = (k+(KA)/(CL))(T-T_(A)) .... (iii)`
Let at `t= 3t_(1)`, temperature of X becomes `T_(2)`
Therefore , from Equation (iii)
`underset(T_(1))overset(T_(2))int(dT)/(T-T_(A))= -(k + (KA)/(LC)) underset(t_(1))overset(3t_(1))intdt`
`implies ln((T_(2) -300)/(350-300))=-2ln(2)-(2KAt_(1))/(LC)`
`implies kt_(1) =ln2` from Equation (i)
This equations gives `T_(2) = (300+ 12.5e^((-2KAt_(1))/(LC)))` kelvin