Find the total heat in the bar given the following initial temperature distribution

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Consider a fully insulated (both laterally and on the ends) bar of length L, that has been heated to an initial temperature distribution of $-x(x-L)$. Find the total heat in the bar for all time.

Above is my problem statement. Here is what I have gathered:

  • The bar is fully insulated, so not heat can enter nor escape the system, other than the initial temperature.
  • The initial temperature distribution implies zero degrees on the ends of the bar and maximum temperature in the centre.
  • Putting the two together, I assume that the bar will eventually come to a non-zero uniform, constant temperature.

So, am I correct to say that the total heat is simply the integral of $-x(x-L)$ for $0$ to $L$, or am I missing something?

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HINT: Let $\rho$ be the linear mass density & $c$ be the specific heat of the bar then consider a bar element of length $dx$

The total heat in the bar $$=\int_{0}^{L} c\ dm T_x=\int_{0}^{L} c(\rho dx) T_x=\rho c\int_{0}^{L}T_x\ dx$$ Put the value of the temperature distribution function $T_x$

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Your bullet points look good. Heat and temperature are different physical quantities, though.

We're given a temperature distribution $T(x)=-x(x-L)$. Integrating temperature for one dimension of length gives you units of $temperature \times length$. That doesn't quite match up to heat, which has units of $energy$.

Heat, $Q$, is related to temperature as $Q=mc T$ where $m$ is the mass and $c$ is the specific heat of the material. Mass depends on the length, $m=\lambda L$ where for a rod, $\lambda$ is the mass density per unit length. The density could be a constant but doesn't have to be, so in general $\lambda = \lambda (x)$. Specific heat $c$ also needn't be a constant, so $c=c(x)$. In general then,

$$Q= mcT=\int_0^L \lambda (x)c(x)T(x) dx= \int_0^L \lambda c\big(-x(x-L)\ \big) dx$$