Certainly! Let's break down the problem step by step to understand the calculations involved.
1. Calculate the heat gained by the water:
The formula for calculating the heat gained (or lost) by a substance is given by:
[tex]\[
q = m \cdot c \cdot \Delta T
\][/tex]
where:
- [tex]\( q \)[/tex] is the heat gained or lost,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
For the water:
- Mass, [tex]\( m = 900 \, \text{g} \)[/tex],
- Specific heat capacity, [tex]\( c = 4.18 \, \text{J/g}^\circ\text{C} \)[/tex],
- Temperature change, [tex]\( \Delta T = 8.32^\circ\text{C} \)[/tex].
Substituting these values into the formula, we get:
[tex]\[
q_{\text{water}} = 900 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 8.32^\circ\text{C}
\][/tex]
The heat gained by the water is:
[tex]\[
q_{\text{water}} = 31299.84 \, \text{J}
\][/tex]
2. Calculate the heat gained by the calorimeter itself:
The calorimeter's heat capacity is given by [tex]\( 2240 \, \text{J/}^\circ\text{C} \)[/tex]. The heat gained by the calorimeter can be calculated using:
[tex]\[
q_{\text{calorimeter}} = C_{\text{calorimeter}} \times \Delta T
\][/tex]
where:
- [tex]\( C_{\text{calorimeter}} \)[/tex] is the heat capacity of the calorimeter,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
Substituting the given values, we get:
[tex]\[
q_{\text{calorimeter}} = 2240 \, \text{J/}^\circ\text{C} \times 8.32^\circ\text{C}
\][/tex]
The heat gained by the calorimeter is:
[tex]\[
q_{\text{calorimeter}} = 18636.8 \, \text{J}
\][/tex]
3. Total heat gained by the calorimeter (water + the calorimeter itself):
The total heat gained is the sum of the heat gained by the water and the calorimeter:
[tex]\[
q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}}
\][/tex]
Substituting the calculated values:
[tex]\[
q_{\text{total}} = 31299.84 \, \text{J} + 18636.8 \, \text{J}
\][/tex]
The total heat gained by the calorimeter is:
[tex]\[
q_{\text{total}} = 49936.64 \, \text{J}
\][/tex]
Therefore, the heat of the calorimeter (water and dry combined) is:
[tex]\[
q_{\text{cal}} = +49936.64 \, \text{J}
\][/tex]
