Rod heated by wire

JEE Advanced 2019 Paper 1, Question 3

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

(1)   \begin{equation*}  T(t) = T_0 (1 + \beta t^{\frac{1}{4}}) \end{equation*}

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,

  1. \frac{4 P [T(t) - T_0]^3}{\beta^4 T_0^4}
  2. \frac{4 P [T(t) - T_0]^4}{\beta^4 T_0^5}
  3. \frac{4 P [T(t) - T_0]^2}{\beta^4 T_0^3}
  4. \frac{4 P [T(t) - T_0]}{\beta^4 T_0^2}


The heat capacity of an object is defined by the relation

(2)   \begin{equation*}  \Delta Q = C \Delta T \end{equation*}

where \Delta Q is the heat that the object absorbs and \Delta T is the resulting temperature change of that object. If the power P delivered to the object is the rate of heat absorption and it is a constant according to the problem. That is,

(3)   \begin{equation*}  P = \frac{dQ}{dt} = {\rm const.} \end{equation*}

Dividing (3) by a small time interval \Delta t and taking the limit \Delta t \to 0 we find

(4)   \begin{align*} \nonumber \lim_{\Delta t \to 0} C \frac{\Delta T}{\Delta t} &= \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t}\\[1em] \implies C \frac{dT}{dt} &= \frac{dQ}{dt} \equiv P \end{align*}

P is a constant and \frac{dT}{dt} = \frac{\beta T_0}{4} t^{-\frac{3}{4}} by (1). That means the heat capacity must be a function of time given by

(5)   \begin{equation*} C(t) = \frac{4P}{\beta T_0} t^{\frac{3}{4}} \end{equation*}

All that remains is to get rid of the time variable in the RHS. We can do this by reorganizing (1) to get

(6)   \begin{equation*} t = \left( \frac{T(t) - T_0}{\beta T_0} \right)^4 \end{equation*}

which can be substitituted into the previous expression to get the final answer

(7)   \begin{equation*} C(t) = \frac{4P [T(t) - T_0]^3}{\beta^4 T_0^4} \end{equation*}

which is option (A).

Leave a Reply