A resistor at a temperature T has a fluctuating voltage. This is a consequence of the fluctuation dissipation theorem which you can use to calculate the spectrum of the voltage. The wikipedia article on the Fluctuation Dissipation Theorem has a section on resistor thermal noise. When measured over a bandwidth $\Delta\nu$, the average squared voltage is:
$$\langle V^2\rangle = 4Rk_BT\Delta\nu$$
where $k_B= 1.38\times 10^{-23}$J/K is Boltzmann's constant.
Suppose we have a resistor $R$ maintained at a temperature T, and hooked up to a resistor $R_0$ initially at absolute zero. The thermal noise of the warm resistor will be as given above. When you apply a voltage $V$ to a resistor $R_0$, the dissipation (in watts) will be given by $IV = V^2R_0$, so the watts applied to the resistor initially at absolute zero, over a bandwidth $\Delta\nu$ will be
$$\langle V^2\rangle = 4R_0Rk_BT\Delta\nu.$$
As that resistor warms up to temperature $T_0$, it will apply a fluctuating voltage on the warm resistor. Following the above, but with the two resistors swapped, the power applied to the warm resistor by the colder resistor at temperature $T_0$ will be:
$$\langle V^2\rangle = 4RR_0k_BT_0\Delta\nu.$$
The system will be in balance when the above two powers are equal. This happens algebraically when $T=T_0$.
A possible source of paradoxical confusion is that the above calculation was done over a limited bandwidth range. But the calculation does not depend on frequency; instead the power transmitted is simply proportional to the range of bandwidths.
For the usual physical system, we consider frequencies that run from 0 to infinity. Thus the total bandwidth is infinite. This suggests that the power flow in the above should be infinite. This paradox is avoided by noting that physical resistors have a limited bandwidth. There is always a parasitic capacitance so that the bandwidth is limited on the high side. Thus the power transfer rate depends on how ideal your resistors are.
As an example calculation, suppose that a resistor has a maximum frequency of 100 GHz $= 10^{11}$ Hz, a (room) temperature of 300K, and a resistance of 1000 ohms. Then the power transfer rate is:
$$ 4\times 1000 \times 1.38\times 10^{-23} 10^{11} \times 300 = 1.66\;\;\textrm{uWatts}$$
Given the heat capacity of the resistor, you can compute the relaxation time with which the colder resistor exponentially approaches an equal temperature.
I think your question is perfectly fine, I don't think this forum is only for advanced research level questions.
Assuming you are not actually receiving small droplets of water, the air around the droplets is cooled by the water, which will then cool your skin. This is strongly accentuated by the fact the water create air currents.
While the effect you suggest of feeling an absence of radiating heat is physically sound, my intuition would tell me it is negligible in this case. Radiation of objects at room temperature wouldn't be very strong compared to the diffusive and convection effects induced by your cold shower.
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It sounds as if your mug is made from Pyrex or some similar low expansion glass. Experiment suggests that if I pour boiling water at 100C into a similar mug in my kitchen (around 15C first thing in the morning) the abrupt 85C temperature change doesn't shatter it. Given this, the 45C difference between room temperature and -25C isn't likely to cause problems, especially since the cooling to -25C will be slow.
You might want to avoid putting the mug down on a very cold surface (which would cause more rapid cooling) but I doubt if even this would have much effect.