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

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Quantrex Academy · Accepted Answer

Correct answer: D. 4P (T(t) - T _ 0 ) ^ 3 β ^ 4 T _ 0 ^ 4

Heat required for small increment in temperature of a substance is given by

dQ = m . s . dT

where,

m =

mass of substances
or

s =

specific heat of substance
Now, heat capacity,

H = m . s .

∴ dQ = H . dT

⇒

Rate of heat absorption by the rod,

\dfrac{dQ}{dt} = P = H ⋅ \dfrac{dT}{dt}

.......(1)
But given that, temperature of rod as a function of time is

T = {T}_{0}\left(1 + \beta{t}^{1 / 4}\right)

......(2)

⇒ \dfrac{dT}{dt} = {T}_{0} ⋅ \beta ⋅ \dfrac{1}{4}{t}^{- \dfrac{3}{4}}

Putting it in (1) gives

P = H ⋅ {T}_{0} ⋅ \beta ⋅ \dfrac{1}{4} ⋅ {t}^{- \dfrac{3}{4}}

⇒ H = \dfrac{4P ⋅ {t}^{3 / 4}}{{T}_{0} ⋅ \beta}

⇒ H = 4P\dfrac{{\left(T - {T}_{0}\right)}^{3}}{{T}_{0}^{4} ⋅ {B}^{4}}

Now using equation

( 2 )

T - {T}_{0} = {T}_{0}\beta ⋅ {t}^{1 / 4}

⇒ {t}^{3 / 4} = {T}_{0}{\left(\dfrac{T - {T}_{0}}{{T}_{0}\beta}\right)}^{3}

A current carrying wire heats a metal rod. The wire provides a...

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