previous iit question

in cooling temp.decreases with time and accord. to newtons law of cooling; dQdt=B(θ-θ₀)=-Cdθdt

-Cdθdt=B(θ-300)

integrate it from (i.e dθ)400 to θ and time from 0 to t

we get
θ=300+ 100exp(-BC)t

and as for t=t₁,θ=350,B={C log_e2/t₁}

now when the body X is connected to a large body Y at atmospheric temp . through a conducting rod of length L and cross sectional area A and conductivity K, the X will lose heat by both radiation and conduction

so dQdt=(dQdt)_C+(dQdt)_R

-Cdθdt=KA(θ-θ₀)/L+B(θ-θ₀)

dθdt=-(θ-300)[KA/CL+ log_e2/t₁]

dθ/(θ-300)=-Zdt where Z=[KA/CL+ log_e2/t₁]

integrate left from 350 to θ_f
and right from t₁to 3t₁

u will get θ_f=300+50exp[-2t₁{KA/CL+ log_e2/t₁]

At t=0; T₀=400K and at t=t₁ ; T₁=350K T_a=350k

Acoording to Newton's law of cooling,

-\frac{dT}{dt}= K(T-T_a)

\Rightarrow \int_{T_{0}}^{T_{1}}{\frac{dT}{T-T_{a}}} =-k\int_{0}^{t_{1}}{dt}

\Rightarrow \ln (\frac{T_{1}-T_{a}}{T_{0}-T_{a}})=-Kt_{1}

\Rightarrow Kt_{1}= -\ln (\frac{350-400}{400-300})=\ln 2

For the second part; At t=3t, Temp=T₂
But here both radiation and conduction takes place.
So,

-\frac{dT}{dt}=K(T-T_{a})+\frac{KA}{CL}(T-T_{a})
Here C= heat capacity of body

\Rightarrow \int_{T_{1}}^{T_{2}}{\frac{dT}{T-T_{a}}}=-(K+\frac{KA}{CL})\int_{t_{1}}^{3t_{1}}{dt}

\Rightarrow \ln (\frac{T_{2}-T_{a}}{T_{1}-T_{a}})=-(2Kt_{1}+\frac{2KAt_{1}}{CL})

\Rightarrow \ln (\frac{T_{2}-300}{50})=-2ln2-\frac{2KAt_{1}}{CL}

\Rightarrow T_{2}=300+\frac{25}{2}e^{-\frac{2KAt_{1}}{CL}} kelvin