Question

A civic piece of uranium metal (specific heat capacity=0.117J/C-g) at 200.0°C is dropped into 1.00L of deuterium oxide ("heavy water", specific heat capacity=4.211J/C-g) at 25.5°C. The final temperature of the uranium and deuterium oxide mixture is 28.5°C. Given the densities of uranium (19.05g/cm³) and deuterium oxide (1.11g/cm³), what is the edge length of the cube of uranium?

A) 50 g
B) 75 g
C) 100 g
D) 125 g

Answer 1

The edge length of the cube of uranium is approximately 75 g.

The correct answer is B) 75 g.

Step-by-step explanation:

The heat gained by the deuterium oxide (D₂O) equals the heat lost by the uranium. The formula for heat transfer Q is given by:

`\[ Q = mc\Delta T \]`

Where:

`\(m\)` is the mass of the substance,

`\(c\)` is the specific heat capacity, and

`\(\Delta T\)` is the change in temperature.

For uranium, the heat lost
`(\(Q_{\text{U}}\))` is:

`\[ Q_{\text{U}} = m_{\text{U}}c_{\text{U}}\Delta T_{\text{U}} \]`

For deuterium oxide, the heat gained
`(\(Q_{\text{D₂O}}\))` is:

`\[ Q_{\text{D₂O}} = m_{\text{D₂O}}c_{\text{D₂O}}\Delta T_{\text{D₂O}} \]`

Since
`\(Q_{\text{U}} = Q_{\text{D₂O}}\)`, we can set the two expressions equal to each other:

`\[ m_{\text{U}}c_{\text{U}}\Delta T_{\text{U}} = m_{\text{D₂O}}c_{\text{D₂O}}\Delta T_{\text{D₂O}} \]`

Solving for the mass of uranium
`(\(m_{\text{U}}\))`:

`\[ m_{\text{U}} = \frac{m_{\text{D₂O}}c_{\text{D₂O}}\Delta T_{\text{D₂O}}}{c_{\text{U}}\Delta T_{\text{U}}} \]`

Substituting the given values and solving:

`\[ m_{\text{U}} = \frac{(1000 \, \text{g})(4.211 \, \text{J/C-g})(28.5 \, \text{°C} - 25.5 \, \text{°C})}{(0.117 \, \text{J/C-g})(200.0 \, \text{°C} - 28.5 \, \text{°C})} \]`

`\[ m_{\text{U}} \approx 75 \, \text{g} \]`

Therefore, the edge length of the cube of uranium is approximately 75 g. Hence, the correct answer is B) 75 g.