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Chapter 6: Problem 32

Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(A\) is larger than that of \(B\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Under the same heating conditions, metal B will take less time to reach a temperature of \(21^{\circ} \mathrm{C}\) compared to metal A, because it has a smaller specific heat capacity.

Step by step solution

01

Recall the definition of specific heat capacity

Specific heat capacity is a property of a substance that measures the amount of heat energy required to increase the temperature of a unit mass of that substance by one degree. A larger specific heat capacity means a substance requires more heat energy to change its temperature.
02

Analysis of heating condition

Now, since both metals A and B are heated under the same conditions (same mass, same initial temperature, and same amount of heat supplied) and we are to find out which metal reaches \(21^{\circ} \mathrm{C}\) first, it is important to consider the effect of their differing specific heats.
03

Determine which metal heats up faster

Because metal A has a larger specific heat, it will take more time to change its temperature by a certain degree compared to metal B. This is because the more specific heat a substance has, the more heat energy it requires to change its temperature. Therefore, metal B would reach a temperature of \(21^{\circ} \mathrm{C}\) before metal A.

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Most popular questions from this chapter

Portable hot packs are available for skiers and people engaged in other outdoor activities in a cold climate. The air-permeable paper packet contains a mixture of powdered iron, sodium chloride, and other components, all moistened by a little water. The exothermic reaction that produces the heat is a very common one- the rusting of iron: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ When the outside plastic envelope is removed, \(\mathrm{O}_{2}\) molecules penetrate the paper, causing the reaction to begin. A typical packet contains \(250 \mathrm{~g}\) of iron to warm your hands or feet for up to \(4 \mathrm{~h}\). How much heat (in \(\mathrm{kJ}\) ) is produced by this reaction? (Hint: See Appendix 2 for \(\Delta H_{\mathrm{f}}^{\circ}\) values.

A quantity of 0.020 mole of a gas initially at \(0.050 \mathrm{~L}\) and \(20^{\circ} \mathrm{C}\) undergoes a constant-temperature expansion until its volume is \(0.50 \mathrm{~L}\). Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (c) If the gas in (b) is allowed to expand unchecked until its pressure is equal to the external pressure, what would its final volume be before it stopped expanding, and what would be the work done?

Define these terms: enthalpy, enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$
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