Heat capacity $C$ is by definition the limit $C=\lim_{\Delta t\to 0}\frac{\Delta Q}{\Delta t}$. Here $\Delta Q$ is the amount of energy transferred as heat onto the material, and $\Delta t$ is the change in temperature. Suppose hot water is placed inside a calorimeter. We wait for the temperatures/thermal energies to settle. We assume that all the thermal energy from the water is absorbed by the calorimeter, that is $\Delta Q_w = \Delta Q_c$.

We have that $$C_{calorimeter}=\lim_{\Delta t \to 0}\frac{\Delta Q_w}{\Delta t_c}=\lim_{\Delta t\to 0}\frac{m_w \cdot C_w\cdot \Delta t_w}{\Delta t_c}$$

But actually, this is $$C_c=\frac{m_w \cdot C_w\cdot \Delta t_w}{\Delta t_c}$$

the equation by which the calculation is done. The inder $w$ denotes the appropriate values for water, $c$ for calorimeter. How can I derive the last equation, when the definition is given with a limit?

  • $\begingroup$ My guess is that the exercise silently assumes $C_w$ (and $m_w$) do not change with temperature - then the equations would hold exactly. This is a reasonable assumption under most practical circumstances (isobaric near-room temperature operation), but it should be stated explicitely. $\endgroup$
    – dominecf
    Dec 28, 2020 at 10:46
  • $\begingroup$ The equation assumes that, over the temperature range of interest, C is constant. $\endgroup$ Dec 28, 2020 at 13:03

2 Answers 2


How can I derive the last equation, when the definition is given with a limit?

Make two basic assumptions:

  1. the process is adiabatic: no external heat enters or leaves the system
  2. the heat capacities are constant over the relevant temperature intervals

The heat balance is thus, with the $\text{LHS}$ the initial heat content and $\text{RHS}$ the final heat content (due to assumption $1$ both are equal to each other):


$$\Rightarrow m_wc_wT_w-m_wc_wT_e=CT_e-CT_c$$

$$\boxed{C=m_w c_w\frac{T_w-T_e}{T_e-T_c}}$$


I've always seen the heat capacity as a proportional coefficient between $\Delta T$, the temperature change, and $Q$, the energy input from heating: $$ Q = C \Delta T$$ as far as the process involves only heat transfer and temperature variation.

I would even go further and say that $C$ is well defined here only for a monophasic body, without any chemical reactions involved and for a constant amount of material $n$.


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.