Question

14. A piece of titanium at 100.0°C was dropped into 50.0 g of water at 20.0°C. The final temperature of the system was 22.6°C. What is the mass of the titanium? (Specific Heat of titanium = 0.54 J/g°C)

Answer 1

Answer:

$m_{Ti}=13.0g$

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:

$-Q_{Ti}=Q_{H_2O}$

That in terms of mass, specific heat and temperatures is:

$-m_{Ti}Cp_{Ti}(T_2-T_{Ti})=m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})$

In such a way, for computing the mass of titanium, considering the heat capacity of water 4.18 J/g°C, we have:

$m_{Ti}=\frac{m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})}{-Cp_{Ti}(T_2-T_{Ti})} \\\\m_{Ti}=\frac{50.0g*4.18\frac{J}{g\°C}(22.6-20.0)\°C}{-0.54\frac{J}{g\°C}*(22.6-100.0)\°C} \\\\m_{Ti}=13.0g$

Regards.