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JEE Advanced 2019 · Paper 1, Physics question 3

A current carrying wire heats a metal rod. The wire provides a constant power (P)(P) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T)(T) in the metal rod changes with time (t)(t) as

T(t)=T0(1+βt14),T(t) = T_0\left(1 + \beta t^{\frac14}\right),

where β\beta is a constant with appropriate dimension while T0T_0 is a constant with dimension of temperature. The heat capacity of the metal is,

Choose one option
Answer and worked solution

Official answer: A

All the power goes into heating the rod: P=CdTdtP=C\dfrac{dT}{dt}.

Rate of heating.
dTdt=βT04t−3/4.\frac{dT}{dt}=\frac{\beta T_0}4t^{-3/4}.

Eliminate tt. From t1/4=T−T0βT0t^{1/4}=\dfrac{T-T_0}{\beta T_0}, we get t−3/4=(βT0T−T0)3t^{-3/4}=\left(\dfrac{\beta T_0}{T-T_0}\right)^3.

Heat capacity.
C=4PβT0(T−T0βT0)3=4P(T−T0)3β4T04.C=\frac{4P}{\beta T_0}\left(\frac{T-T_0}{\beta T_0}\right)^3=\frac{4P(T-T_0)^3}{\beta^4T_0^4}.
(A)

Topics:Calorimetry and specific heatDimensional analysis
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