3in 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 loge2/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+ loge2/t1]
dθ/(θ-300)=-Zdt where Z=[KA/CL+ loge2/t1]
integrate left from 350 to θf
and right from t1to 3t1
u will get θf=300+50exp[-2t1{KA/CL+ loge2/t1]
1At t=0; T0=400K and at t=t1 ; T1=350K Ta=350k
Acoording to Newton's law of cooling,
-\frac{dT}{dt}= K(T-Ta)
\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
1
varun.tinkle
·2010-06-14 05:06:35
yeah thnx i was also doing by this method but the calculus blew me off to Mars