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][tex]$(g)$[/tex][/tex] & 200.0 \\
\hline Specific heat [tex][tex]$\left(J / g ^{\circ}C \right)$[/tex][/tex] & 4.18 \\
\hline [tex][tex]$T_{i} \left(^{\circ}C \right)$[/tex][/tex] & 20.1 \\
\hline [tex][tex]$T_{f} \left(^{\circ}C \right)$[/tex][/tex] & 45.1 \\
\hline
\end{tabular}

Calculate the heat absorbed ([tex][tex]$J$[/tex][/tex]).



Answer :

To determine the amount of energy absorbed by the water in the experiment using the calorimeter data, follow these steps:

1. Identify the given values:
- Mass of water ([tex]\( m \)[/tex]): 200.0 grams
- Specific heat capacity of water ([tex]\( c \)[/tex]): 4.18 J/g°C
- Initial temperature ([tex]\( T_i \)[/tex]): 20.1°C
- Final temperature ([tex]\( T_f \)[/tex]): 45.1°C

2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
The change in temperature is given by the difference between the final and initial temperatures:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substituting the given values:
[tex]\[ \Delta T = 45.1°C - 20.1°C = 25.0°C \][/tex]

3. Apply the formula for heat absorbed ([tex]\( Q \)[/tex]):
The formula for the heat absorbed by the water is:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat absorbed (in Joules)
- [tex]\( m \)[/tex] is the mass of the water
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature

4. Substitute the values into the formula:
[tex]\[ Q = 200.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times 25.0°C \][/tex]

5. Calculate the heat absorbed:
[tex]\[ Q = 200.0 \times 4.18 \times 25.0 = 20900.0 \, \text{J} \][/tex]

Therefore, the amount of energy absorbed by the water in the experiment is [tex]\( 20900.0 \)[/tex] Joules.