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Chapter 6: Problem 126

Ice at \(0^{\circ} \mathrm{C}\) is placed in a Styrofoam cup containing \(361 \mathrm{~g}\) of a soft drink at \(23^{\circ} \mathrm{C}\). The specific heat of the drink is about the same as that of water. Some ice remains after the ice and soft drink reach an equilibrium temperature of \(0^{\circ} \mathrm{C}\). Determine the mass of ice that has melted. Ignore the heat capacity of the cup.

Short Answer

Expert verified
Approximately 104 grams of ice has melted

Step by step solution

01

Calculate the heat lost by the drink

First, calculate the heat lost by the drink as it cools down to \(0^{\circ} \mathrm{C}\). This can be found using the formula \(q = mcΔT\), where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since the specific heat capacity of the drink is approximately the same as water, use \(c = 4.184 \mathrm{J/g^{\circ}C}\). Plug in the given values to find \(q = 361\, \mathrm{g} \times 4.184\, \mathrm{J/g^{\circ}C} \times (23^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) = 34727.744\, \mathrm{J}\)
02

Calculate the mass of ice that has melted

The heat gained by the ice equals the heat lost by the drink, and this heat goes into melting the ice. Use the heat of fusion of ice, which is \(334\, \mathrm{J/g}\), to find the mass of the ice that has melted. Given that the heat is equal to the mass times the heat of fusion, you can rearrange to find mass = heat / heat of fusion = \(34727.744\, \mathrm{J} / 334 \, \mathrm{J/g} = 104.046\, \mathrm{g}\)

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Most popular questions from this chapter

Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? (The specific heats of water vapor and air are approximately \(1.9 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C},\) respec- tively.)

Which of the following does not have \(\Delta H_{\mathrm{f}}^{\circ}=0\) at \(25^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { He }(g) & \text { Fe }(s) & \text { Cl }(g) & \text { S }_{8}(s) & \text { O }_{2}(g) & \text { Br }_{2}(l)\end{array}\)

From the following heats of combustion, $$\begin{array}{r}\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}(\mathrm{graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow\mathrm{CH}_{3} \mathrm{OH}(l)$$

From the enthalpy of formation for \(\mathrm{CO}_{2}\) and the following information, calculate the standard enthalpy of formation for carbon monoxide (CO). $$ \begin{aligned} \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H^{\circ} &=-283.0 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$Why can't we obtain it directly by measuring the enthalpy of the following reaction?$$\mathrm{C}(\text { graphite })+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)$$

Which is the more negative quantity at \(25^{\circ} \mathrm{C}: \Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g) ?\)
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