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

Question

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:
$T (t) = T_{0} (1 + β t^{1 / 4})$
where $β$ is a constant with appropriate dimension while $T_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is:

TestHub · Accepted Answer

Heat required for small increment in temperature of a substance is given by
$d Q = m . s . d T$
where, $m =$ mass of substances
or $s =$ specific heat of substance
Now, heat capacity, $H = m . s .$
$∴ d Q = H . d T$
$\Rightarrow$ Rate of heat absorption by the rod, $\frac{d Q}{d t} = P = H \cdot \frac{d T}{d t}$ .......(1)
But given that, temperature of rod as a function of time is
$T = T_{0} (1 + β t^{1 / 4})$ ......(2)
$\Rightarrow \frac{d T}{d t} = T_{0} \cdot β \cdot \frac{1}{4} t^{- \frac{3}{4}}$
Putting it in (1) gives
$P = H \cdot T_{0} \cdot β \cdot \frac{1}{4} \cdot t^{- \frac{3}{4}}$
$\Rightarrow H = \frac{4 P \cdot t^{3 / 4}}{T_{0} \cdot β}$
$\Rightarrow H = 4 P \frac{{(T - T_{0})}^{3}}{T_{0}^{4} \cdot B^{4}}$
Now using equation $(2)$
$T - T_{0} = T_{0} β \cdot t^{1 / 4}$
$\Rightarrow t^{3 / 4} = T_{0} {(\frac{T - T_{0}}{T_{0} β})}^{3}$

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