Answer:
[tex]C_{zinc}=0.400\frac{J}{g\°C}[/tex]
Explanation:
Hello!
In this case, by considering thermodynamics and the temperature of zinc, we can infer it is hot while water is cold, it means that the heat lost by the zinc is gained by the water and we can write:
[tex]Q_{Zn}=-Q_{water}[/tex]
In terms of mass, specific heat and temperatures we can write:
[tex]m_{Zn}C_{Zn}(T_{eq}-T_{Zn})=-m_{water}C_{water}(T_{eq}-T_{water})[/tex]
In such a way, by solving for the specific heat capacity of the zinc we write:
[tex]C_{zinc}=\frac{-m_{water}C_{water}(T_{eq}-T_{water})}{m_{Zn}(T_{eq}-T_{Zn})}[/tex]
Thus, by plugging in the given values we obtain:
[tex]C_{zinc}=\frac{-45.0g*4.184\frac{J}{g\°C}* (27.1\°C-98.8\°C)}{13.8g*(27.1\°C-98.8\°C)}\\\\C_{zinc}=0.400\frac{J}{g\°C}[/tex]
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