Let's solve the problem step by step.
### Given Data
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): [tex]\(2.00 \, \text{kg}\)[/tex]
2. Temperature change ([tex]\(\Delta T\)[/tex]): [tex]\(0.658 \, ^\circ \text{C}\)[/tex]
3. Heat capacity of the dry portion of the calorimeter ([tex]\(C_{\text{dry}}\)[/tex]): [tex]\(1050 \, \text{J/(ºC)}\)[/tex]
4. Specific heat capacity of water ([tex]\(C_{\text{H}_2\text{O}}\)[/tex]): [tex]\(4184 \, \text{J/(kg} \cdot \,^ \circ \text{C)}\)[/tex]
### Steps to Solve
1. Calculate the heat absorbed by the water ([tex]\(q_{\text{H}_2\text{O}}\)[/tex]):
[tex]\[
q_{\text{H}_2\text{O}} = m_{\text{water}} \times C_{\text{H}_2\text{O}} \times \Delta T
\][/tex]
Substituting in the values:
[tex]\[
q_{\text{H}_2\text{O}} = 2.00 \, \text{kg} \times 4184 \, \text{J/(kg} \cdot \,^ \circ \text{C)} \times 0.658\, ^\circ \text{C}
\][/tex]
[tex]\[
q_{\text{H}_2\text{O}} = 5506.144 \, \text{J}
\][/tex]
2. Calculate the heat absorbed by the dry portion of the calorimeter ([tex]\(q_{\text{dry}}\)[/tex]):
[tex]\[
q_{\text{dry}} = C_{\text{dry}} \times \Delta T
\][/tex]
Substituting in the values:
[tex]\[
q_{\text{dry}} = 1050 \, \text{J/(ºC)} \times 0.658\, ^\circ \text{C}
\][/tex]
[tex]\[
q_{\text{dry}} = 690.9 \, \text{J}
\][/tex]
3. Calculate the total heat absorbed by the calorimeter ([tex]\(q_{\text{cal}}\)[/tex]):
[tex]\[
q_{\text{cal}} = q_{\text{H}_2\text{O}} + q_{\text{dry}}
\][/tex]
Substituting in the values:
[tex]\[
q_{\text{cal}} = 5506.144 \, \text{J} + 690.9 \, \text{J}
\][/tex]
[tex]\[
q_{\text{cal}} = 6197.044 \, \text{J}
\][/tex]
### Conclusion
The heat of the overall calorimeter is [tex]\(\boxed{6197.044 \, \text{J}}\)[/tex].
