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Chapter 7: Problem 4

A 75.0 g piece of \(\mathrm{Ag}\) metal is heated to \(80.0^{\circ} \mathrm{C}\) and dropped into \(50.0 \mathrm{g}\) of water at \(23.2^{\circ} \mathrm{C} .\) The final temperature of the \(\mathrm{Ag}-\mathrm{H}_{2} \mathrm{O}\) mixture is \(27.6^{\circ} \mathrm{C}\). What is the specific heat of silver?

Short Answer

Expert verified
The specific heat of silver is \(0.235 J/g°C\).

Step by step solution

01

Calculate the Heat Gained by Water

First, let's calculate the heat gained by water using the formula \( q = mc\Delta T \). Here the mass \( m \) is \(50.0 g\), \( c \) (the specific heat capacity of water) is \( 4.18 J/g°C \), and \(\Delta T \) (the change in temperature) is \( 27.6°C - 23.2° C = 4.4° C \). Substituting these values into the equation, we find \( q = mc\Delta T = 50.0 g \times 4.18 J/g°C \times 4.4 °C = 921.92 J \). This is the heat gained by the water.
02

Calculate the Heat Lost by silver

Next, understand that the heat gained by the water is equal to the heat lost by silver. Thus, the heat lost by the silver, which we'll denote by \( q_{Ag} \), is also \(921.92 J\).
03

Calculate the Specific Heat of Silver

Finally, knowing the heat lost by the silver, you can calculate the specific heat of silver using the same formula \( q = mc\Delta T \). This time you know \( q_{Ag} = 921.92 J\), \( m = 75.0 g \), and \( \Delta T = 80°C - 27.6°C = 52.4°C \). Plugging these values in the equation and solving for \( c \), you get: \( c = q / m\Delta T = 921.92 J / (75.0 g \times 52.4°C ) = 0.235 J/g°C \). The specific heat of silver is therefore \( 0.235 J/g°C \).

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Most popular questions from this chapter

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)

Hot water and a piece of cold metal come into contact in an isolated container. When the final temperature of the metal and water are identical, is the total energy change in this process (a) zero; (b) negative; (c) positive; (d) not enough information.

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