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Why do we mix water with different temperatures to calculated the heat capacity of the calorimeter? Why can't we calculate it directly using just warm water?

Asked differently, would the following be wrong? (With $T_E$ the equilibrium temperature and $T_C$ and $T_W$ the initial temperatures of the caloriemeter and water)

$$ c_{C}=\frac{\Delta Q}{m_{C} \Delta T}=\frac{c_{W} m_{W}\left(T_{E}-T_{W}\right)}{m_{C}\left(T_{C}-T_{E}\right)} $$

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  • $\begingroup$ You need a temperature range to calculate heat exchange (Recall $Q_\text{lost or gain} = m \times c \times \Delta T$ where $c$ is specific heat of the solution). $\endgroup$
    – Mathew Mahindaratne
    Commented Jun 18, 2020 at 19:31
  • $\begingroup$ Wouldn't it be sufficient to know the initial temperature of the caloriemeter, the water and the resulting temperature of both? $\endgroup$
    – klugreuter
    Commented Jun 18, 2020 at 19:52
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    $\begingroup$ In a thought experiment, a lot of things are possible. To do it for real, you need a reproducible experiment with a step-by-step list of actions. $\endgroup$
    – Karl
    Commented Jun 18, 2020 at 20:56
  • $\begingroup$ @klugreuter: Where is the specific heat capacity of the caloriemeter (assuming $m_C$ is the mass of caloriemeter)? And, now you realize their should be a temperature range. $\endgroup$
    – Mathew Mahindaratne
    Commented Jun 18, 2020 at 21:55
  • $\begingroup$ The specific heat capacity of the caloriemeter is the thing im trying to determine, named $c_C$ above. Sorry, my knowledge about thermodynamics and chemistry is extremely limited, I'm just trying to figure out why - when we want to get some information about how much energy is lost to the system - we cant just put in some energy and see how much is lost afterwards. $\endgroup$
    – klugreuter
    Commented Jun 18, 2020 at 22:08

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Karl's comment (see above) prompt me to suggest the following rough experiment which can be done in the classroom with minimum equipment:

Suppose you are required to find out the specific heat capacity of water. Weigh $m_1 \ \pu{g}$ of warm water at $T_1 \ \pu{^\circ C}$ to a Styrofoam cup with a lid. Also weigh $m_2 \ \pu{g}$ of ice at $T_2 \ \pu{^\circ C}$ to another Styrofoam cup. Then add ice to warm water (Note: Measure the temperature of warm water just before addition to make sure it is still at $T_1 \ \pu{^\circ C}$). Stir the mixture carefully while the coup is covered by the lid. Measure the temperature of the mixture until reading is steady. Suppose the steady temperature is $T_3 \ \pu{^\circ C}$. Ensure this time no ice is visible (meaning all dissolved). Now we are ready for the calculations:

Suppose the specific heat capacity of water is $c_{(\ell)}$ and the specific heat capacity of water is $c_{(s)}$. For novices, we can assume that $c_{(\ell)} \approx c_{(s)} = c$. Thus, heat lost by warm water at equilibrium:

$$Q_\text{lost} = m_1 \times c \times (T_1 - T_3) \tag1$$

For ice, if $T_2 \lt 0$, ice would gain heat first to become ice at $\pu{0 ^\circ C}$, then $\pu{0 ^\circ C}$ ice melts to $\pu{0 ^\circ C}$ water, and warm up to $T_3 \ \pu{^\circ C}$ water. If heat of fusion of ice is $L \ \pu{Jg-1}$, total heat gained by ice at equilibrium:

$$Q_\text{gained} = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) \tag2$$

At equilibrium, $Q_\text{gained} = Q_\text{lost}$. Therefore, equations $(1)$ and $(2)$:

$$m_1 \times c \times (T_1 - T_3) = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) tag3$$

Only unknown in the equation $(3)$ is $c$, you can solve the equation $(3)$ for $c$.

Also note that you can avoid $m_2 \times c \times (0 - T_2)$ part by choosing $\pu{0 ^\circ C}$ ice, because it is easy to get $\pu{-20 ^\circ C}$ from freezer, it it kept in constant temperature.

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