How much energy does the water in this experiment absorb according to the calorimeter data?

\begin{tabular}{|l|l|}
\hline \multicolumn{2}{|c|}{ Calorimeter Data } \\
\hline Mass [tex]$(g)$[/tex] & 100.0 \\
\hline Specific heat [tex]$\left(J/g^{\circ}C\right)$[/tex] & 4.18 \\
\hline [tex]$T_{i}\left(^{\circ}C\right)$[/tex] & 21.2 \\
\hline [tex]$T_{f}\left(^{\circ}C\right)$[/tex] & 46.2 \\
\hline
\end{tabular}

[tex]$q_{H_{2}O} = [?] \text{J}$[/tex]



Answer :

To find out how much energy the water absorbed in the experiment according to the calorimeter data, we'll follow these steps:

1. Determine the given values from the calorimeter data:
- Mass of the water, [tex]\( m = 100.0 \, \text{g} \)[/tex]
- Specific heat of water, [tex]\( c = 4.18 \, \text{J/g}^\circ\text{C} \)[/tex]
- Initial temperature, [tex]\( T_i = 21.2^\circ\text{C} \)[/tex]
- Final temperature, [tex]\( T_f = 46.2^\circ\text{C} \)[/tex]

2. Calculate the change in temperature [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substituting the values:
[tex]\[ \Delta T = 46.2^\circ\text{C} - 21.2^\circ\text{C} = 25.0^\circ\text{C} \][/tex]

3. Use the formula for the amount of heat absorbed or released in a calorimetry process:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]
where [tex]\( q \)[/tex] is the heat absorbed, [tex]\( m \)[/tex] is the mass, [tex]\( c \)[/tex] is the specific heat, and [tex]\(\Delta T\)[/tex] is the change in temperature.

4. Substitute the values into the formula:
[tex]\[ q = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 25.0^\circ\text{C} \][/tex]

5. Calculate the energy absorbed:
[tex]\[ q = 100.0 \times 4.18 \times 25.0 = 10450.0 \, \text{J} \][/tex]

Therefore, the energy the water absorbed in this experiment is [tex]\( 10450.0 \, \text{J} \)[/tex].