A current carrying wire heats a metal rod. The wire provides a constant power to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature of the metal rod changes with time as [2019]

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 $(T)$ of the metal rod changes with time $(t)$ as $T (t) = T_{0} (1 + β t^{2 / 4})$

Where $β$ is a constant with appropriate dimensions while $T_{0}$ is a constant with dimensions of temperature. The heat capacity of metal is with: [2019]

Smart Achievers · Accepted Answer

(4)

$Power,$

$P = \frac{d Q}{d t} = \frac{d}{d t} (m c) T = (H) \frac{d T}{d t} = (H) \frac{d}{d t} [T_{0} (1 + β t^{1 / 4})]$

$P = (H) T_{0} = \frac{β t^{- 3 / 4}}{4} where H = heat capacity$

$∴$ $(H) = \frac{4 P t^{3 / 4}}{T_{0} β} \dots (i)$

But $t^{1 / 4} = \frac{T (t) - T_{0}}{β T_{0}}$

$∴$ $t^{3 / 4} = \frac{{[T (t) - T_{0}]}^{3}}{β^{3} T_{0}^{3}} \dots (i i)$

From eq. (i) & (ii),

$(H) = \frac{4 P}{T_{0} β} \cdot \frac{{[T (t) - T_{0}]}^{3}}{β^{3} T_{0}^{3}} = \frac{4 P {[T (t) - T_{0}]}^{3}}{β^{4} T_{0}^{4}}$

Solution

1 $\frac{4 P (T (t) - T_{0})}{β^{4} T_{0}^{2}}$

2 $\frac{4 P (T {(t) - T_{0})}^{2}}{β^{4} T_{0}^{3}}$

3 $\frac{4 P (T {(t) - T_{0})}^{4}}{β^{4} T_{0}^{5}}$

4 $\frac{4 P (T {(t) - T_{0})}^{3}}{β^{4} T_{0}^{4}}$

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Solution

1 4P(T(t)-T0)β4T02

2 4P(T(t)-T0)2β4T03

3 4P(T(t)-T0)4β4T05

4 4P(T(t)-T0)3β4T04

Ans. (4) Power, P=dQdt=ddt(mc)T=(H)dTdt=(H)ddt[T0(1+βt1/4)] P=(H)T0=βt-3/44 where H=heat capacity ∴ (H)=4Pt3/4T0β ⋯(i) But t1/4=T(t)-T0βT0 ∴ t3/4=[T(t)-T0]3β3T03 ⋯(ii) From eq. (i) & (ii), (H)=4PT0β·[T(t)-T0]3β3T03=4P[T(t)-T0]3β4T04

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1 $\frac{4 P (T (t) - T_{0})}{β^{4} T_{0}^{2}}$

2 $\frac{4 P (T {(t) - T_{0})}^{2}}{β^{4} T_{0}^{3}}$

3 $\frac{4 P (T {(t) - T_{0})}^{4}}{β^{4} T_{0}^{5}}$

4 $\frac{4 P (T {(t) - T_{0})}^{3}}{β^{4} T_{0}^{4}}$