Question

A cylindrical block of length $0.4$ m and area of cross-section $0.04m^2$ 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 $10$ watt/m-K and the specific heat of the material of the disc is $600$ J/kg-K, it would take $X\times 83secondsforthetemperatureofthedisctoincreaseto$350$ K? Assume ,for calculation purpose, 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.The value of x is:

Answer 1

Let $\theta$ be the temperature of disc at any time $t$. The quantity of heat conducted by cylinder to disc in time $dt$ is given by

$d\theta =kA\frac{[{\theta }_o-\theta ]}{l}dt\left(1\right)$

( l - length of cylinder )

This quantity of heat $d\theta$ increases the temperature of disc from $\theta$ to $d\theta$, therefore

$dQ=msd\theta \left(2\right)$

m - mass

s - specific heat of disc

Equating (1) and (2)

$kA\frac{[{\theta }_o-\theta ]}{l}dt=msd\theta \Rightarrow dt=\frac{msl}{kA}\left(\frac{d\theta }{{\theta }_o-\theta }\right)$

To calculate the time in raising the temperature of disc from $300K$ to $350K$:

${\int }_0^{t}dt={\int }_{300}^{350}\frac{msl}{kA}\left(\frac{d\theta }{{\theta }_o-\theta }\right)\Rightarrow t=\frac{msl}{kA}{\int }_{300}^{350}\frac{d\theta }{{\theta }_o-\theta }\Rightarrow t=\frac{msl}{kA}{[-\log ({\theta }_o-\theta )]}_{300}^{350}=\frac{msl}{kA}{\log }_e\left(\frac{{\theta }_o-300}{{\theta }_o-350}\right)\Rightarrow t=\frac{0.4\times 600\times 0.4}{10\times 0.04}{\log }_e\left(\frac{400-300}{400-350}\right)=240{\log }_{e}2\Rightarrow t=240\times 0.6931=166.3s\approx 166sec$