Q. how to calculate heat capacity of calorimeter

Step 1. State the physical setup and the quantity to find. You mix a known mass of hot water with a known mass of cold water inside the calorimeter, wait until thermal equilibrium, and measure the final equilibrium temperature. The calorimeter has an unknown heat capacity, which we denote by \(C_{\text{cal}}\). The specific heat capacity of water is \(c_w\). The known quantities are the hot water mass \(m_h\), its initial temperature \(T_h\), the cold water mass \(m_c\), its initial temperature \(T_c\), and the final equilibrium temperature \(T_f\). The sign convention used is that heat lost by a hotter body is positive when we compute the magnitude lost, and heat gained by colder bodies is positive when we compute the magnitude gained. Energy conservation requires that heat lost by the hot water equals heat gained by the cold water plus heat gained by the calorimeter.

Step 2. Write the energy-balance equation. The heat lost by the hot water is \(m_h c_w (T_h – T_f)\). The heat gained by the cold water is \(m_c c_w (T_f – T_c)\). The heat gained by the calorimeter is \(C_{\text{cal}} (T_f – T_c)\) if the calorimeter starts at temperature \(T_c\) (the same as the cold water). Energy conservation gives

\[
m_h c_w (T_h – T_f) = m_c c_w (T_f – T_c) + C_{\text{cal}} (T_f – T_c).
\]

Step 3. Solve this equation for the unknown calorimeter heat capacity \(C_{\text{cal}}\). First isolate the term containing \(C_{\text{cal}}\).

\[
C_{\text{cal}} (T_f – T_c) = m_h c_w (T_h – T_f) – m_c c_w (T_f – T_c).
\]

Now divide both sides by \((T_f – T_c)\) to obtain \(C_{\text{cal}}\).

\[
C_{\text{cal}} = \frac{m_h c_w (T_h – T_f) – m_c c_w (T_f – T_c)}{T_f – T_c}.
\]

Step 4. Simplify the formula by factoring out \(c_w\) if desired.

\[
C_{\text{cal}} = c_w \,\frac{m_h (T_h – T_f) – m_c (T_f – T_c)}{T_f – T_c}.
\]

Step 5. Example numeric calculation. Suppose \(c_w = 4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}\), \(m_h = 0.200\ \text{kg}\), \(T_h = 80.0\ ^\circ\text{C}\), \(m_c = 0.150\ \text{kg}\), \(T_c = 20.0\ ^\circ\text{C}\), and \(T_f = 48.0\ ^\circ\text{C}\). Compute each heat term first.

\[
\text{Heat lost by hot water} = m_h c_w (T_h – T_f) = 0.200\times 4186\times (80.0 – 48.0) = 26\,790.4\ \text{J}.
\]

\[
\text{Heat gained by cold water} = m_c c_w (T_f – T_c) = 0.150\times 4186\times (48.0 – 20.0) = 17\,581.2\ \text{J}.
\]

\[
\text{Difference} = 26\,790.4 – 17\,581.2 = 9\,209.2\ \text{J}.
\]

\[
T_f – T_c = 48.0 – 20.0 = 28.0\ \text{K}.
\]

\[
C_{\text{cal}} = \frac{9\,209.2}{28.0} = 328.9\ \text{J}\,\text{K}^{-1}\quad\text{(approximately)}.
\]

Step 6. Units and interpretation. The result \(C_{\text{cal}}\) has units of joules per kelvin, J K^{-1}. It represents the amount of heat required to raise the calorimeter temperature by one kelvin. In the example above, the calorimeter requires about \(329\ \text{J}\) to raise its temperature by \(1\ ^\circ\text{C}\).

Step 7. Alternate method using electrical heating. If instead you supply a known electrical energy \(Q_{\text{elec}}\) to the calorimeter (for example by a heater) and measure the temperature rise \(\Delta T\), then the combined heat capacity of calorimeter plus any water inside is

\[
C_{\text{total}} = \frac{Q_{\text{elec}}}{\Delta T}.
\]

If you know the mass of water \(m_w\) present, you can separate the calorimeter heat capacity as

\[
C_{\text{cal}} = \frac{Q_{\text{elec}}}{\Delta T} – m_w c_w.
\]

Step 8. Notes on experimental practice and uncertainties. Measure masses and temperatures carefully. Use the same temperature scale so that temperature differences are in kelvin. Account for heat losses to the environment if they are significant, by insulating the calorimeter or applying corrections. Propagate measurement uncertainties through the formula to obtain uncertainty in \(C_{\text{cal}}\).