Heat and Thermodynamics 3 Question 37
37. A cylindrical block of length $0.4 m$ and area of cross-section $0.04 m^{2}$ is placed coaxially on a thin metal disc of mass $0.4 kg$ and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of $400 K$ and the initial temperature of the disc is $300 K$. If the thermal conductivity of the material of the cylinder is $10 W / mK$ and the specific heat capacity of the material of the disc is $600 J / kg-K$, how long will it take for the temperature of the disc to increase to $350 K$ ? Assume, for purposes of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.
(1992, 8M)
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Answer:
Correct Answer: 37. $166.32 s$
Solution:
- Let at any time temperature of the disc be $\theta$. At this moment rate of heat flow,
$$ \frac{d Q}{d t}=\frac{K A(\Delta \theta)}{l}=\frac{K A}{l}\left(\theta _0-\theta\right) $$
This heat is utilised in increasing the temperature of the disc. Hence,
$$ \frac{d Q}{d t}=m s \frac{d \theta}{d t} $$
Equating Eqs. (i) and (ii), we have
$$ m s \frac{d \theta}{d t}=\frac{K A}{l}\left(\theta _0-\theta\right) $$
Therefore, $\quad \frac{d \theta}{\theta _0-\theta}=\frac{K A}{m s l} d t$
or
$$ \int _{300 K}^{350 K} \frac{d \theta}{\theta _0-\theta}=\frac{K A}{m s l} \int _0^{t} d t $$
or
$$ \therefore \quad t=\frac{m s l}{K A} \ln \left(\frac{\theta _0-300}{\theta _0-350}\right) $$
Substituting the values, we have
$$ \begin{aligned} & t=\frac{(0.4)(600)(0.4)}{(10)(0.04)} \ln \left(\frac{400-300}{400-350}\right) \\ & t=166.32 s \end{aligned} $$
