An unknown volume of water at 18.2°C is added to 24.4 mL

Step 1 of 3

Here, we are going to determine the unknown volume of water at \(23.5^{\circ} \mathrm{C}\).

The heat capacity is directly related to the temperature change.

                                                   \(q=c m \Delta T\)

Where \(c\) is the heat capacity, \(q\) is the heat absorbed by the object \(m\) is the mass of the object

\(\Delta T\) is the change in temperature \(=\left(T_{\text {final }}-T_{\text {initial }}\right)\)

Therefore,

                                                    \(q=c m \Delta T\)

                                              \(\Rightarrow\left(T_{\text {final }}-T_{\text {initial }}\right)=\frac{q}{c m} \ldots(1)\)