Answer:
Explanation:
Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):
[tex]DH_{lead}=-DH_{water}[/tex]
Now, by stating the heat capacity definition:
[tex]m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\[/tex]
Solving for the equilibrium temperature:
[tex]T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC[/tex]
Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.
Best regards.
