/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 We can determine the purity of s... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Look up the specific heat of several elements, and plot the products of the specific heats and atomic masses as a function of the atomic masses. Based on the plot, develop a hypothesis to explain the data. How could you test your hypothesis?

Construct a concept map encompassing the ideas behind the first law of thermodynamics.

A 74.8 g sample of copper at \(143.2^{\circ} \mathrm{C}\) is added to an insulated vessel containing \(165 \mathrm{mL}\) of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}(\mathrm{l})(d=1.26 \mathrm{g} / \mathrm{mL}),\) at \(24.8^{\circ} \mathrm{C} .\) The final temperature is \(31.1^{\circ} \mathrm{C}\). The specific heat of copper is \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) What is the heat capacity of glycerol in \(\mathrm{Jmol}^{-1}\) \(^{\circ} \mathrm{C}^{-1} ?\)

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

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