Answer:
[tex]Cp_{metal}=39.0\frac{J}{g\°C}[/tex]
Explanation:
Hello,
In this case, since the following equation relates heat, mass, specific heat and temperature:
[tex]Q=mCp(T_2-T_1)[/tex]
For the two substances, we say that the heat lost by the metal equals the heat gained by the water, thus we have:
[tex]m_{water}Cp_{water}(T_2-T_{water})=-m_{metal}Cp_{metal}(T_2-T_{metal})[/tex]
Thus, the specific heat of the metal turns out:
[tex]Cp_{metal}=\frac{m_{water}Cp_{water}(T_2-T_{water})}{m_{metal}(T_2-T_{metal})}[/tex]
Therefore, we obtain:
[tex]Cp_{metal}=\frac{125g*4.18\frac{J}{g\°C} *(31.0-91.0)\°C}{-134.0g*(31.0-25.0)\°C}\\\\Cp_{metal}=39.0\frac{J}{g\°C}[/tex]
Best regards.
