Question

A cubic piece of uranium metal (specific heat capacity \(=0.117\) \(\mathrm{J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(200.0^{\circ} \mathrm{C}\) is dropped into \(1.00 \mathrm{~L}\) deuterium oxide ("heavy water," specific heat capacity \(=4.211 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(25.5^{\circ} \mathrm{C}\). The final temperature of the uranium and deuterium oxide mixture is \(28.5^{\circ} \mathrm{C}\). Given the densities of uranium \(\left(19.05 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and deuterium oxide (1.11 \(\mathrm{g} / \mathrm{mL}\) ), what is the edge length of the cube of uranium?

Short answer

The edge length of the uranium cube is approximately 2.64 cm.

Step-by-step solution

Find the mass of deuterium oxide

Since the volume of the deuterium oxide is given in liters, we need to convert it to mL so that the volume will be compatible with the density which is given in \(\mathrm{g/mL}\). \[ 1.00 \,\mathrm{L} = 1.00 \times 1000\,\mathrm{mL} = 1000\,\mathrm{mL} \] Now that we have the volume of deuterium oxide in mL, we can use the density of deuterium oxide to find its mass. \[ mass_{D_2O} = \text{volume}_{D_2O} \times density_{D_2O} \] \[ mass_{D_2O} = 1000\,\mathrm{mL} \times 1.11\,\mathrm{g/mL} = 1110\,\mathrm{g} \]

Equate the heat gained by deuterium oxide to the heat lost by uranium

Heat gained by deuterium oxide is given by: \( Q_{D_2O} = m_{D_2O} \times c_{D_2O} \times \Delta T_{D_2O} \) Heat lost by uranium is given by: \( Q_U = m_U \times c_U \times \Delta T_U \) Since the heat gained by deuterium oxide is equal to the heat lost by uranium, we can write: \( m_{D_2O} \times c_{D_2O} \times \Delta T_{D_2O} = m_U \times c_U \times \Delta T_U \) Now, let's plug in the given values for specific heat capacities and temperatures: \( 1110\,\mathrm{g} \times 4.211\,\mathrm{\frac{J}{^\circ C \cdot g}} \times (28.5 - 25.5)^\circ \mathrm{C} = m_U \times 0.117\,\mathrm{\frac{J}{^\circ C \cdot g}} \times (200.0 - 28.5)^\circ \mathrm{C} \)

Solve for the mass of uranium

Now we can solve for the mass of uranium, \(m_U\): \( m_U = \frac{1110\,\mathrm{g} \times 4.211\,\mathrm{\frac{J}{^\circ C \cdot g}} \times (28.5 - 25.5)^\circ \mathrm{C}}{0.117\,\mathrm{\frac{J}{^\circ C \cdot g}} \times (200.0 - 28.5)^\circ \mathrm{C}} \) \( m_U \approx 349.53\,\mathrm{g} \)

Calculate the volume of uranium

Now that we have the mass of uranium, we can find its volume using its density: \[ \text{volume}_U = \frac{mass_U}{density_U} \] \[ \text{volume}_U = \frac{349.53\,\mathrm{g}}{19.05\,\mathrm{g/cm^3}} \approx 18.34\,\mathrm{cm^3} \]

Find the edge length of the uranium cube

Since the uranium is in the shape of a cube, we can find the edge length by taking the cube root of its volume: \[ edge\text{ }length_U = \sqrt[3]{\text{volume}_U} \] \[ edge\text{ }length_U = \sqrt[3]{18.34\,\mathrm{cm^3}} \approx 2.64\,\mathrm{cm} \] The edge length of the uranium cube is approximately 2.64 cm.

Key concepts

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). In simpler terms, it indicates the thermal inertia of a material—the higher the specific heat capacity, the more energy it takes to change the temperature of a given amount of the substance.

This concept is crucial because it explains why different materials heat up or cool down at different rates when they absorb or lose the same amount of heat. For example, in the provided exercise, uranium has a lower specific heat capacity \(0.117 \frac{J}{^\circ C \cdot g}\) compared to deuterium oxide (heavy water), which has a specific heat capacity of \(4.211 \frac{J}{^\circ C \cdot g}\). This means that for each gram, uranium's temperature will change more rapidly than that of deuterium oxide for the same amount of heat transfer.

In the context of the textbook exercise, understanding specific heat is essential for determining how much the temperature of the uranium and deuterium oxide will change upon mixing and reaching a thermal equilibrium.

Heat Transfer Calculations

Heat transfer calculations are crucial in thermochemistry to determine the amount of heat required to change a substance's temperature or to find out the final temperature when two substances at different temperatures come into contact. The key formula used for such calculations is the equation \( Q = mc\Delta T \), where \(^`Q^`\) is the heat transfer, \(^`m^`\) is the mass of the substance, \(^`c^`\) represents specific heat capacity, and \(^`\Delta T^`\) is the change in temperature (final temperature - initial temperature).

Applying the Heat Transfer Formula

When the cubic piece of uranium and deuterium oxide mix, heat will flow from the hotter uranium to the cooler deuterium oxide until both reach the same temperature, which is the final temperature of 28.5°C. By equating the heat lost by uranium to the heat gained by deuterium oxide, one can solve for either a change in temperature or for an unknown mass, as done in the step-by-step solution provided.

Using this equation helps ensure that the principle of conservation of energy is honored, as energy cannot be created or destroyed, only transferred.

Density and Volume Relationships

The density of a substance is defined as its mass per unit volume, often expressed as \( \frac{g}{cm^3} \) or \( \frac{g}{mL} \). Density is a fundamental property of matter and is particularly important when one needs to convert between mass and volume. As density is mass divided by volume (\( \rho = \frac{m}{V} \)), knowing two of these values allows you to calculate the third.

Understanding Density in the Context of the Exercise

In the exercise, the known densities of uranium \( (19.05 \frac{g}{cm^3}) \) and deuterium oxide \( (1.11 \frac{g}{mL}) \) allow for the calculation of their respective masses from given volumes. Then, the volume of uranium could be further calculated from its mass and known density. This is vital for solving the final part of the problem – determining the dimension of the uranium cube.

To find the edge length of the uranium cube, the volume is calculated from the mass and density, and then the cubic root of the volume gives the length of the edge of the cube. These concepts show the interplay of mass, volume, and density in practical applications, such as determining the size of a substance based on its physical properties.