Heat Transfer.[doubt]

Let at some later time t, the temperatures of the two bodies be T₁'(t) and T₂'(t) respectively. So that the temperature difference is ΔT(t) = T₁'(t) &ndash T₂'(t) as a function of time. The heat current at this instant is given by
\dot{q}=kA\dfrac{\Delta T}{L} ------ (*)
In the next infinitesimal interval of time δt, the heat current transfers \dot{q}\ \mathrm \delta t from first body to the second one. So the first body looses this amount of energy because of which its temperature drops to T₁'(t+δt) while the second body gains this much of energy so its temperature increases to T₂'(t+δt). So the new temperature difference is
ΔT(t+δt) = T₁'(t+δt) &ndash T₂'(t+δt)
So we have
ΔT(t+δt) &ndash ΔT(t) = (T₁'(t+δt) &ndash T₁'(t)) &ndash ( T₂'(t+δt) &ndash T₂'(t)) ------- (1)
But \dot{q}\ \mathrm \delta t = &ndash m₁ s₁ (T₁'(t+δt) &ndash T₁'(t)) = m₂ s₂( T₂'(t+δt) &ndash T₂'(t))
Hence, (1) becomes
ΔT(t+δt) &ndash ΔT(t) = &ndash \dot{q}\ \mathrm \delta tm₁s₁ &ndash \dot{q}\ \mathrm \delta tm₂ s₂
from where we obtain
&ndash ΔT(t+δt) &ndash ΔT(t)δt =\dot{q}\left(\dfrac{1}{m_1s_1}+\dfrac{1}{m_2s_2}\right)
In the limit δt →0, we get
\dot q = -\dfrac{1}{C}\dfrac{\mathrm d(\Delta T)}{\mathrm dt}
where
C=\dfrac{1}{m_1s_2}+\dfrac{1}{m_2s_2}
Substituting this in (*), we get
-\dfrac{1}{C}\dfrac{\mathrm d(\Delta T)}{\mathrm dt}=kA\dfrac{\Delta T}{L}
which could be integrated to give
\boxed{\Delta T =(T_1-T_2)e^{-t/\tau}}
where
\dfrac{1}{\tau}=\dfrac{kA}{L}\left(\dfrac{1}{m_1s_1}+\dfrac{1}{m_2s_2}\right)

ya that i think is correct

thanks [1]

now
dT_{1}=\frac{KA\Delta T}{m_{1}s_{1}L} dt
similarly
dT_{2}=\frac{KA\Delta T}{m_{2}s_{2}L} dt
dT=\frac{KA\Delta T}{m_{1}s_{1}L} dt+\frac{KA\Delta T}{m_{2}s_{2}L} dt
now solve this by integrating it (here \Delta T is variable )
(i think there is a Sign convention error )

let at time t the temperature difference is \Delta T
now heat transfer in very small time is
dQ=\frac{KA\Delta T}{L}d t
now heat losed by one part is = heat gained by other part
dQ=m_{1}s_{1}dT_{1}=m_{2}s_{2}dT_{2}
where dT(1) and dT(2) are small change in temp of m(1) and m(2)

now the total small temperature difference is dT=dT_{1}+dT_{2}

i will complete the soln in next post

@prince can u jus post the complete soln

my ans isnt matching

yar just use haet current principle that heat passis through the both bodies are sme( heat current=κA TEMP DIFF/d) let the temp of junction any

[2]

ya its an hcv q

plz somebody post the soln

???

help neone...plzzzz

its an HCV question

ya i got till der

but jus unable to do teh integrating part.....i mean ans isnt matching

we shud do lik this na

dT/dt=the exp which u got in teh last step

can u plz poast a complete soln....

seriously i m not able to get it [2]

plz help

no one ????? [2]