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

adiabatic calorimeter
(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.

adiabatic calorimeter with two bodies


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.