# Temperature variation after a magnetization change using magnetic entropy?

I'm doing some theoretical work on magnetism and anisotropic magnetic entropy change, and I'm stuck at an early step in my calculation. The system I'm studying is similar to the experimental set-up of specific heat, minus the heating element (the source of power $$P(t)$$). I've started my math here for a leaking system (heat capacity calculus) :
$$C \frac{d T}{d t} = P(t)- K_b (T - T_0)$$ with the $$P(t)=0$$:
$$C \frac{d T}{d t} = - K_b (T - T_0)$$ Using the second law of thermodynamics, which states (which I believe I remember wrong):
$$C_{\beta,\gamma,...}^{(\alpha)} = T \frac{d S_{\beta,\gamma,...} }{d \alpha}$$ Since specific heat is an extensive (and so is entropy) variable:
$$C = C_L + C_M = C_L + T \frac{d S_M}{d T}$$ I can rewritte the second equation as

$$C_L \frac{d T}{d t} = - T \frac{d S_M}{d T} \frac{d T}{d t} - K_b (T- T_0)$$ Here is the root of my problem. I've done the following, assuming a relation between entropy and its orientation $$\theta$$:

$$\frac{d T}{d t} = \frac{1}{C_L} \left\lbrack -T \frac{d S_M}{d T} \frac{dT}{dt} \frac{d \theta}{d \theta} - K_b (T - T_0) \right\rbrack$$ $$= \frac{1}{C_L} \left\lbrack -T \frac{d S_M}{d \theta} \frac{d \theta}{d t} - K_b (T - T_0) \right\rbrack$$

Which I doubt can be done so (mathematically at least). I'm there stuck on theoretical details, wondering if I can even do math properly, or if there's a subtility to the thing that I missed. Is it correct or is it wrong?

Thank you all for your time, cheers.