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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 \((19.05\) \(\mathrm{g} / \mathrm{cm}^{3}\) ) and deuterium oxide \((1.11 \mathrm{~g} / \mathrm{mL})\), what is the edge length of the cube of uranium?

Short Answer

Expert verified
The edge length of the uranium cube is approximately 1.316 cm, which is found by equating the heat transfers of uranium and deuterium oxide, then calculating the mass of the uranium cube, and finally finding the volume and edge length of the cube.

Step by step solution

01

Write the heat transfer equation

Since the final temperature of the uranium and deuterium oxide mixture is given, we will assume that no energy is lost to the environment and the equilibrium is reached when the heat gained by deuterium oxide is equal to the heat lost by uranium metal. We can write this as: Q_uranium = -Q_deuterium_oxide
02

Calculate the mass of deuterium oxide

We are given the volume of deuterium oxide as 1.00 L, and we know its density (1.11 g/mL). We can calculate its mass using the following equation: mass_deuterium_oxide = volume × density mass_deuterium_oxide = 1.00 L × (1.11 g/mL) × (1000 mL/L) = 1110 g
03

Write heat transfer equations for uranium and deuterium oxide

Now, we can write the heat transfer equations for both uranium and deuterium oxide using the formula Q = mcΔT, where m is mass, c is specific heat capacity and ΔT is the change in temperature. For uranium: Q_uranium = mass_uranium × c_uranium × (T_final - T_initial_uranium) For deuterium oxide: Q_deuterium_oxide = mass_deuterium_oxide × c_deuterium_oxide × (T_final - T_initial_deuterium_oxide)
04

Equate the heat transfer equations

Next, we can equate the heat transfer equations as per step 1: mass_uranium × c_uranium × (T_final - T_initial_uranium) = -mass_deuterium_oxide × c_deuterium_oxide × (T_final - T_initial_deuterium_oxide)
05

Calculate mass of the uranium cube

Now, we can plug in the known values and solve for the mass of the uranium cube: mass_uranium × (0.117 J/g°C) × (28.5°C - 200°C) = - (1110 g) × (4.211 J/g°C) × (28.5°C – 25.5°C) mass_uranium ≈ 42.95 g
06

Calculate the volume of the uranium cube

Since we know the density of uranium (19.05 g/cm³), we can calculate the volume of the uranium cube as follows: volume_uranium = mass_uranium / density_uranium volume_uranium ≈ 42.95 g / (19.05 g/cm³) ≈ 2.255 cm³
07

Calculate the edge length of the cube

Finally, we can now find the edge length of the cube since we know its volume. For a cube, we know that volume = edge_length³. So, edge_length = (volume_uranium)^(1/3) edge_length ≈ (2.255 cm³)^(1/3) ≈ 1.316 cm The edge length of the uranium cube is approximately 1.316 cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Equation
Understanding the heat transfer equation is essential for thermal equilibrium calculations. This is the principle that allows us to calculate how much energy is transferred as heat between objects of different temperatures. The key formula we use is Q = mcΔT, where Q represents the heat transferred, m is the mass of the substance, c stands for the specific heat capacity of the substance, and ΔT (delta T) is the change in temperature.

In our exercise, heat is transferred between the uranium cube and deuterium oxide until thermal equilibrium is reached - the point where they both reach the same final temperature. No heat is lost to the surroundings, meaning the heat lost by the uranium cube (Q_uranium) equals the heat gained by the deuterium oxide (-Q_deuterium_oxide). This allows students to set one heat transfer equation equal to the negative of the other and solve for unknown quantities. This exercise also highlights the convention that heat lost is considered negative, a crucial concept in thermodynamics.

For a more comprehensive understanding, it's key to discuss the conservation of energy and how it applies to closed systems like our scenario - thereby strengthening the conceptual foundation beyond just memorizing equations.
Specific Heat Capacity
The specific heat capacity (c) is a property of matter that indicates how much heat energy (in joules or calories) is needed to raise the temperature of one gram of the substance by one degree Celsius (or Kelvin). It's a measure of a substance's 'thermal inertia' - how resistant it is to temperature change.

In practice, substances with high specific heat capacities can absorb a lot of heat without a large change in temperature. Water, for instance, has a high specific heat capacity, making it effective for cooling systems. In our exercise, the deuterium oxide has a higher specific heat capacity than uranium, which plays a crucial role in how much heat the uranium can transfer and how this affects the temperature of the system.

Emphasizing the implications of specific heat capacity, such as why land near the ocean has a more moderate climate compared to inland areas, can help students attach real-world meaning to this scientific concept.
Density and Volume Calculations
Density is a fundamental property that relates the mass of a substance to its volume, often expressed as grams per cubic centimeter (g/cm³) for solids and liquids. The formula used for calculating density is density = mass/volume. To find volume when the mass and density are known, we rearrange this equation to volume = mass/density.

Understanding how to manipulate this equation is essential in science and engineering, particularly when designing objects that need to achieve a specific weight or buoyancy level. In our exercise, we calculate the uranium cube's volume using its density and mass, and then find the edge length, assuming the shape is a perfect cube. Real-world applications such as determining the amount of material needed for a product or the load an object can carry in water can be directly connected to these calculations.

Breaking down composite shapes into simpler geometric volumes or discussing the impact of temperature on density can expand upon this topic, demonstrating its versatility in problem-solving across various disciplines.

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Most popular questions from this chapter

Are the following processes exothermic or endothermic? a. the combustion of gasoline in a car engine b. water condensing on a cold pipe c. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) d. \(\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g)\)

Explain why oceanfront areas generally have smaller temperature fluctuations than inland areas.

On Easter Sunday, April 3,1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\left.\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\right)\) b. According to The Denver Post for April 4,1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

Consider an airplane trip from Chicago, Illinois, to Denver, Colorado. List some path-dependent functions and some state functions for the plane trip.

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