Answer:
[tex]m_{Ti}=13.0g[/tex]
Explanation:
Hello,
In this case, based on the given, we can infer that as titanium is hot and water cold, it cools down whereas the water is heated up, therefore, in terms of heat, we have that the heat lost by the titanium is gained by the water:
[tex]-Q_{Ti}=Q_{H_2O}[/tex]
That in terms of mass, specific heat and temperatures is:
[tex]-m_{Ti}Cp_{Ti}(T_2-T_{Ti})=m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})[/tex]
In such a way, for computing the mass of titanium, considering the heat capacity of water 4.18 J/g°C, we have:
[tex]m_{Ti}=\frac{m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})}{-Cp_{Ti}(T_2-T_{Ti})} \\\\m_{Ti}=\frac{50.0g*4.18\frac{J}{g\°C}(22.6-20.0)\°C}{-0.54\frac{J}{g\°C}*(22.6-100.0)\°C} \\\\m_{Ti}=13.0g[/tex]
Regards.
