Newton's Law of Cooling on a calorimeter with two systems

Having the previous calorimeter with adiabatic walls, we've proven, based on Newton's law of cooling, that the equations are...

(source: gyazo.com)

$$m_{b} \times C _{pb} \frac{\mathrm{d T_{b}} }{\mathrm{d} t} = k_{t}(T_{c} - T_{b})$$

$$m_{c} \times C _{pc} \frac{\mathrm{d T_{c}} }{\mathrm{d} t} = k_{t}(T_{b} - T_{c})$$

How would it be affected if instead of a single system, it uses two bodies $$c1$$ and $$c2$$ that never get in touch where $$c1$$ has a temperature bigger than the water and $$c2$$ has a temperature that is smaller than the water.

Just apply the energy balance once again to the water. The rate of accumulation of energy in the water is equal to the rate of heat gain from c1 minus the rate of heat loss to c2:

$$m_wC_{pw} \frac{dT_w}{dt} = hA\left ( T_{c_1} - T_w \right ) - hA\left ( T_w - T_{c_2} \right )$$

Notice I have introduced the area $A$, since $h\Delta T$ is a flux, or heat transfer per unit area.