Magnetism and Thermodynamics

There is no magnetism in this problem, but only the use of the definition of heat capacity and power. Let the temperature varies with time as $T(t)=T_0(1+βt^n)$ where $T_0$ is the initial temperatue and $β,n$ are constants.

Then, by definitions of heat capacity and power, the power $P$ is related to the heat capacity $C$ through :

$P=C\dfrac{dT}{dt}=βnCT_0t^{n-1}$

$\implies C=\dfrac{P}{βnT_0t^{n-1}}$ .

Now, from the expression for $T(t)$ ,

we get $t^{n-1}=\dfrac{(T(t)-T_0)^{1-\frac{1}{n}}}{β^{1-\frac{1}{n}}T_0^{1-\frac{1}{n}}}$ .

So, $C=\dfrac{P\left (T(t)-T_0\right )^{\frac{1}{n}-1}}{nβ^{\frac{1}{n}}T_0^{\frac{1}{n}}}$ .

In this problem, $n=\dfrac{1}{4}$ .

So, $C=\dfrac{P\left (T(t)-T_0\right )^3}{\frac{1}{4}β^4T_0^4}$ .

Therefore $α=4, \phi =3, \gamma =\delta =4$ , and $α+\phi +\gamma +\delta =4+3+4+4=\boxed {15}$ .

Applying the first law of thermodynamics on the rod in terms of power:

$P_{in} = P_{absorbed}$

$P$ stands for power. Let the heat capacity be $C$ . Then:

$P = C \frac{dT}{dt}$ $\implies P = CT_o \beta \frac{t^{-3/4}}{4}$ $\implies C = \frac{4P}{T_o \beta} t^{3/4}$

Consider:

$T= T_o\left(1 + \beta t^{1/4}\right)$ $t^{3/4} = \frac{(T-T_o)^3}{\beta^3 T_o^3}$

Using the above:

$C = \frac{4P}{T_o \beta} t^{3/4}$ $C = \frac{4P}{T_o \beta} \left(\frac{(T-T_o)^3}{\beta^3 T_o^3}\right)$ $C = \frac{4P(T-T_o)^3}{T_o^4 \beta^4}$

$\alpha = 4$ $\phi = 3$ $\gamma = 4$ $\delta = 4$