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A current carrying wire heats a metal rod. The wire provides a constant power $( P )$ to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature ( $T$ ) in the metal rod changes with time $( t )$ as :
$T ( t )= T _0\left(1+\beta t ^{1 / 4}\right)$
where $\beta$ is a constant with appropriate dimension while $T _0$ is a constant with dimension of temperature.
The heat capacity of the metal is :
$\frac{4 P \left( T ( t )- T _0\right)^3}{\beta^4 T _0^4}$
$\frac{4 P \left( T ( t )- T _0\right)}{\beta^4 T _0^2}$
$\frac{4 P \left( T ( t )- T _0\right)^4}{\beta^4 T _0^5}$
$\frac{4 P \left( T ( t )- T _0\right)^2}{\beta^4 T _0^3}$
Solution
$P =\frac{ dQ }{ dt } \quad T _{( t )}= T _0\left(1+\beta t ^{1 / 4}\right)$
$\frac{ dQ }{ dt }= ms \frac{ dT }{ dt } \Rightarrow S =\frac{ P }{\left(\frac{ dT }{ dt }\right)}$
$\frac{d T}{d t}=T_0\left[0+\beta \frac{1}{4} \cdot t ^{-3 / 4}\right]=\frac{\beta T_0}{4} \cdot t ^{-3 / 4}$
$S =\frac{ P }{( dT / dt )}=\frac{4 P }{\beta T _0} \cdot t ^{3 / 4}$
$S=\frac{4 P}{\beta}\left[\frac{ t ^{3 / 4}}{ T _0}\right]$
$\frac{ T ( t )}{ T _0}=\left(1+\beta t ^{1 / 4}\right)$
$\beta t ^{1 / 4}=\frac{ T ( t )}{ T _0}-1=\frac{ T ( t )- T _0}{ T _0}$
$t ^{3 / 4}=\left(\frac{ T ( t )- T _0}{\beta \cdot T _0}\right)^3$
$\Rightarrow S =\frac{4 P }{ T _0 \beta}\left[\frac{ T ( t )- T _0}{\beta \cdot T _0}\right]^3=\frac{4 P }{\beta^4 T _0^4}\left[ T ( t )- T _0\right]^3$
