Answer:
[tex]T_{eq}=23.85^oC[/tex]
Explanation:
Hello,
In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:
[tex]\Delta H_{Cu}=-\Delta H_{H_2O}[/tex]
Therefore the equilibrium temperature shows up as:
[tex]m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\[/tex]
Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:
[tex]T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC[/tex]
Best regards.
