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We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

Short Answer

Expert verified
To conclude, the decision regarding the purity of the gold ring depends on the comparison of the calculated amounts of heat transferred. If the heat lost by the ring coincides with the heat gained by the water, it shows that the ring is pure gold. Otherwise, it is not.

Step by step solution

01

Calculation of heat lost by the gold ring

The heat lost by the gold ring can be calculated using the formula \(q = mc\Delta T\), where \(q\) is the heat transferred, \(m\) is the mass of the substance, \(c\) is the specific heat of the substance, and \(\Delta T\) is the change in temperature. For the gold ring, its mass \(m = 10.5 \, \mathrm{g}\),its specific heat \(c = 0.1291 \, \mathrm{J/g} \degree \mathrm{C}\),and the change in temperature \(\Delta T = 78.3^{\circ} \mathrm{C} - 31^{\circ} \mathrm{C} = 47.3^{\circ} \mathrm{C}\)So plugging these values into the formula gives: \(q_{\text{gold}} = 10.5 \, \mathrm{g} \times 0.1291 \, \mathrm{J/g} \degree \mathrm{C} \times 47.3^{\circ} \mathrm{C}\)
02

Calculation of heat gained by the water

The heat gained by water can also be calculated using the formula \(q = mc\Delta T\). Here,\(m = 50.0 \, \mathrm{g}\) (mass of the water),\(c = 4.18 \, \mathrm{J/g} \degree \mathrm{C}\) (specific heat of the water),and \(\Delta T = 31^{\circ} \mathrm{C} - 23.7^{\circ} \mathrm{C} = 7.3^{\circ} \mathrm{C}\) (change in temperature)We plug these values into the formula to find the heat gained by the water: \(q_{\text{water}} = 50.0 \, \mathrm{g} \times 4.18 \, \mathrm{J/g} \degree \mathrm{C} \times 7.3^{\circ} \mathrm{C}\)
03

Verification

Finally, we verify if the heat lost by the gold equals the heat gained by water (\(q_{\text{gold}} = q_{\text{water}}\)). If this holds true, then the gold ring is pure. Otherwise, it is impure.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental concept in calorimetry. It represents the amount of heat required to change the temperature of a substance by one degree Celsius per unit mass.
In this exercise, we see the specific heat capacity of gold pegged at \(0.1291\, \mathrm{J/g \degree C}\). This suggests that gold requires \(0.1291\, \mathrm{J}\) to increase the temperature of each gram by one degree Celsius.

Understanding specific heat capacity helps us predict how different substances will respond to heat changes. For instance, substances with low specific heat, like metals, heat up and cool down quickly.
Conversely, substances like water, with higher specific heat, absorb more heat before their temperature rises significantly.
  • Gold's low specific heat makes it excellent for jewelry as it stays relatively even in temperature despite being exposed to various thermal conditions.
Heat Transfer Calculations
Heat transfer calculations allow us to understand how heat moves between objects. In this calorimetry problem, heat transfers from the hotter gold ring to the colder water. We calculate this using the formula: \[ q = mc\Delta T \]Where
  • \(q\) is the heat transferred,
  • \(m\) is the mass,
  • \(c\) is the specific heat,
  • \(\Delta T\) is the change in temperature.
For the gold ring, we find the heat lost using
  • \(m = 10.5 \, \mathrm{g}\),
  • \(c = 0.1291 \, \mathrm{J/g \degree C}\),
  • \(\Delta T = 47.3^{\circ} \mathrm{C}\).
And for the water, the heat gained calculations use
  • \(m = 50.0 \, \mathrm{g}\),
  • \(c = 4.18 \, \mathrm{J/g \degree C}\),
  • \(\Delta T = 7.3^{\circ} \mathrm{C}\).
By equating the heat lost by the gold to the heat gained by the water, we test if energy is conserved.
Heat transfer calculations help chemists understand reactions and engineers design thermal systems.
Chemical Purity Analysis
Chemical purity analysis in calorimetry involves determining the purity of a substance based on its thermal properties and reactions. To analyze the purity of the gold ring, the exercise examines if the heat lost by gold equals the heat gained by water.
This equality suggests the absence of impurities, which might have different heat capacities, potentially altering heat exchange values. If there is no discrepancy in these heat exchanges, the gold is likely pure.

Purity analysis ensures the authenticity and quality of materials. In industrial applications, such assessments are crucial for quality control and meeting regulatory standards.
  • Pure materials maintain expected thermal behavior, while impurities can cause unexpected thermal interactions.
By employing calorimetry, professionals can detect and correct these variations effectively. This guarantees that substances like precious metals meet high standards, preserving their intrinsic and market value.

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Most popular questions from this chapter

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

A calorimeter that measures an exothermic heat of reaction by the quantity of ice that can be melted is called an ice calorimeter. Now consider that \(0.100 \mathrm{L}\) of methane gas, \(\mathrm{CH}_{4}(\mathrm{g}),\) at \(25.0^{\circ} \mathrm{C}\) and \(744 \mathrm{mm} \mathrm{Hg}\) is burned at constant pressure in air. The heat liberated is captured and used to melt \(9.53 \mathrm{g}\) ice at \(0^{\circ} \mathrm{C}\left(\Delta H_{\text {fusion }} \text { of ice }=6.01 \mathrm{kJ} / \mathrm{mol}\right)\) (a) Write an equation for the complete combustion of \(\mathrm{CH}_{4},\) and show that combustion is incomplete in this case. (b) Assume that \(\mathrm{CO}(\mathrm{g})\) is produced in the incomplete combustion of \(\mathrm{CH}_{4}\), and represent the combustion as best you can through a single equation with small whole numbers as coefficients. \((\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is another . product of the combustion.)

There are other forms of work besides \(\mathrm{P}-\mathrm{V}\) work. For example, electrical work is defined as the potential \(x\) change in charge, \(w=\phi d q\). If a charge in a system is changed from \(10 \mathrm{C}\) to \(5 \mathrm{C}\) in a potential of \(100 \mathrm{V}\) and \(45 \mathrm{J}\) of heat is liberated, what is the change in the internal energy? (Note: \(1 \mathrm{V}=1 \mathrm{J} / \mathrm{C})\).

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

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