I shall try to explain using first law of thermodynamics and Maxwell' relations. For simplicity I am considering $C_v$ . You may extend the argument to $C_p$ as well. Let us consider the change in $C_v$ with w.r.t $V$. From the definition of $C_v$ we can write,
$\left(\frac{dC_v}{dV}\right)_T = \left[\frac{d}{dV}\left(\frac{dU}{dT}\right)_V\right]_T ...(1)$
Switch order of differentiation and from the first law of thermodynamics we can write
$\left[\frac{d}{dT}\left(\frac{dU}{dV}\right)_T\right]_V = \left[\frac{d}{dT}\left(T \frac{dS}{dV} - P \right)_T\right]_V .. (2)$
From Maxwell's relation we have
$ \left(\frac{dS}{dV}\right)_T = \left(\frac{dP}{dT}\right)_V .. (3) $
Substitute Eqn $(3)$ in $(2)$ and after applying product rule, you may obtain
$\left(\frac{dC_v}{dV}\right)_T = T \left(\frac{d^2P}{dT^2}\right)_V$
For an ideal gas or a gas having an equation of state of the form $PV = kT$ or a linear function of $T$, the second derivative of $P$ w.r.t $T$ vanishes and therefore no dependence. For any other equation of state, this is the dependence of specific heat w.r.t to pressure or volume. You may work out in a similar fashion for $C_p$ w.r.t $P$. Only difference is that the thermodynamic quantity will be Enthalpy and not Internal Energy
If the heat capacity varies with temperature, then the heat balance condition for determining the final equilibrium temperature T in this problem is: $$\int_0^T{C(T')dT'}=\int_T^{100}{C(T')dT'} $$where C(T') is the heat capacity at temperature T'. From the mean value theorem for integration, we know that $$\int_0^T{C(T')dT'}=C(T_1)T$$and $$\int_T^{100}{C(T')dT'}=C(T_2)(T-100)$$where $T_1$ is some temperature between 0 and T, and $T_2$ is some temperature between T and 100. So, we have: $$C(T_1)T=C(T_2)(T-100)$$So, solving for T, we obtain:
$$T=100\frac{C(T_2)}{(C(T_1)+C(T_2)}=50\left[1+\frac{C(T_2)-C(T_1)}{(C(T_1)+C(T_2))}\right]$$
From the problem statement, we know that the heat capacity increases with temperature, so $C(T_2)>C(T_1)$. This means that the second term in brackets is positive, and the final temperature is greater than 50 C.
Best Answer
In general, heat capacity depends on temperature, so the answer is no, the amount is different. However, due to the equipartition theorem, at sufficiently high temperature (compared to the typical temperature scale given by quantum mechanics) the temperature dependence flattens out. Famously, this gives rise to Dulong-Petit's law for solids or the constant heat capacity of ideal gases that you must have seen. This is why to observe temperature dependence you need to go to low temperature (take for example Debye's heat capacity for solids), or include many degrees of freedom (take blackbody radiation). As a consequence of the 3rd law of thermodynamics or more generally a finite value of residual entropy, you also need to have a vanishing of the heat capacity at low temperature.
The archetypical example is the harmonic oscillator. At pulsation $\omega$, you have the following formula (using the canonical ensemble) for $C$, the heat capacity: $$ C = k_B \left(\frac{x}{\sinh x}\right)^2 $$ with $x=\hbar\omega/2k_BT$, which gives $C=k_B$ in the high temperature limit (ie $T\gg \hbar\omega/k_B$) as predicted by the equipartition theorem. However, in the low energy limit, quantum effects dominate, and you have the signature exponential decrease of $C$ due to the energy gap of the groundstate.
Hope this helps and tell me if you need more detail.