Question

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('heavy water,' specific heat capacity = 4.211 J/°C･g) at 25.5°C. The final temperature of the platinum and deuterium oxide mixture is 41.9°C. The density of platinum is 21.45 g/cm³ and the density of deuterium oxide is 1.11 g/mL. What is the edge length of the cube of platinum, in centimeters?

Answer 1

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.