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Chapter 7: Problem 46

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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Most popular questions from this chapter

Brass has a density of \(8.40 \mathrm{g} / \mathrm{cm}^{3}\) and a specific heat of \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} . \mathrm{A} 15.2 \mathrm{cm}^{3}\) piece of brass at an initial temperature of \(163^{\circ} \mathrm{C}\) is dropped into an insulated container with \(150.0 \mathrm{g}\) water initially at \(22.4^{\circ} \mathrm{C}\) What will be the final temperature of the brass-water mixture?

A 125 \(g\) stainless steel ball bearing \((\mathrm{spht}=\) \(0.50 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\) at \(525^{\circ} \mathrm{C}\) is dropped into \(75.0 \mathrm{mL}\) of water at \(28.5^{\circ} \mathrm{C}\) in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches \(100.0^{\circ} \mathrm{C} .\) What mass of water vaporizes while the boiling continues? \(\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

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: \(1 \mathrm{V}=1 \mathrm{J} / \mathrm{C})\).

Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

When one mole of sodium carbonate decahydrate (washing soda) is gently warmed, \(155.3 \mathrm{kJ}\) of heat is absorbed, water vapor is formed, and sodium carbonate heptahydrate remains. On more vigorous heating, the heptahydrate absorbs \(320.1 \mathrm{kJ}\) of heat and loses more water vapor to give the monohydrate. Continued heating gives the anhydrous salt (soda ash) while \(57.3 \mathrm{kJ}\) of heat is absorbed. Calculate \(\Delta H\) for the conversion of one mole of washing soda into soda ash. Estimate \(\Delta U\) for this process. Why is the value of \(\Delta U\) only an estimate?
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