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Chapter 11: Problem 47

A quantity of ice at \(0^{\circ} \mathrm{C}\) is added to \(64.3 \mathrm{~g}\) of water in a glass at \(55^{\circ} \mathrm{C}\). After the ice melted, the temperature of the water in the glass was \(15^{\circ} \mathrm{C}\). How much ice was added? The heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\) and the specific heat is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
32.26 grams of ice were added.

Step by step solution

01

Energy Released by Water Cooling

Calculate the energy released by the water as it cools from \(55^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\). Use the specific heat capacity formula:\[q = m \cdot c \cdot \Delta T\]where \(q\) is the heat energy, \(m\) is the mass of the water (64.3 g), \(c\) is the specific heat capacity (4.18 J/g路掳C), and \(\Delta T\) is the change in temperature (55掳C - 15掳C = 40掳C).First, convert \(c\) to kJ by multiplying by \(0.001\), so it becomes \(0.00418\) kJ/g路掳C.Calculate:\[ q = 64.3 \times 0.00418 \times 40 = 10.75816 \text{ kJ} \]
02

Convert Energy from kJ to Quantities of Ice

Calculate how many moles of ice this energy could melt, using the heat of fusion formula:\[\text{Moles of ice} = \frac{q}{\text{heat of fusion}} = \frac{10.75816 \text{ kJ}}{6.01 \text{ kJ/mol}}\]Calculate:\[\frac{10.75816}{6.01} \approx 1.79 \text{ moles of ice} \]
03

Convert Moles of Ice to Grams

Convert the moles of ice to grams. The molar mass of water is approximately 18.02 g/mol. Calculate:\[\text{Mass of ice} = 1.79 \text{ moles} \times 18.02 \text{ g/mol} = 32.26 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat of Fusion
The concept of "heat of fusion" is crucial when dealing with phase changes like melting. It refers to the amount of energy needed to change 1 mole of a substance from solid to liquid at its melting point, without a change in temperature. For water, the heat of fusion is 6.01 kJ/mol.
This means that 6.01 kJ of energy is required to melt 1 mole of ice at 0掳C. When ice absorbs this energy, it transitions into liquid water. Conversely, when water freezes, it releases the same amount of energy.
The heat of fusion is a key component in calorimetry problems involving ice and water, as it determines how much ice can be melted by a given amount of energy.
Specific Heat Capacity
Specific heat capacity is a measure of how much energy it takes to raise the temperature of 1 gram of a substance by 1掳C. For water, the specific heat capacity is 4.18 J/g路掳C, which is relatively high.
Because of its high specific heat capacity, water can absorb a lot of heat without a large increase in temperature. This property makes water an excellent medium for storing and transferring heat. In the given problem, the energy released by the cooling water is calculated using this property.
The formula used is:
  • \( q = m \cdot c \cdot \Delta T \)
  • \( q \) is the heat energy absorbed or released.
  • \( m \) is the mass of the substance.
  • \( c \) is the specific heat capacity.
  • \( \Delta T \) is the change in temperature (in 掳C).
Molar Mass
Molar mass is the mass of 1 mole of a substance, expressed in grams per mole (g/mol). It is a fundamental concept used in stoichiometry to convert between mass and amount in moles.
For water, the molar mass is approximately 18.02 g/mol. This value is calculated by adding the atomic masses of two hydrogen atoms (1.01 g/mol each) and one oxygen atom (16.00 g/mol).
In the problem, after determining the amount of melted ice in moles, we convert this to grams using the formula:
  • \( \text{Mass of ice} = \text{moles of ice} \times \text{molar mass of water} \)
This conversion is essential for understanding how much ice was originally added.
Energy Conservation
Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transferred or transformed. In calorimetry problems, this principle is applied to calculate energy exchanges between substances.
In the exercise, the principle of energy conservation is observed when the energy lost by warm water is equal to the energy gained by the melting ice. This is why the calculations start with determining the energy released by the water cooling down.
Once the energy lost by the water is known, it is used to find out how much ice can melt using the heat of fusion. This direct energy transfer between the water and the ice, without any loss in the system, illustrates energy conservation in action.

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Most popular questions from this chapter

Consider the following two compounds: \(\begin{array}{ll}\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & \text { 1-pentanol }\end{array}\) \(\begin{array}{ll}\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} & \text { hexane }\end{array}\) What are the different types of intermolecular forces that exist in each compound? One of these compounds has a normal boiling point of \(69^{\circ} \mathrm{C}\), and the other has a normal boiling point of \(138^{\circ} \mathrm{C}\). What is the normal boiling point of hexane? Explain. One of these compounds has a viscosity of \(0.313 \mathrm{~g} /(\mathrm{cm} \cdot \mathrm{s})\), and the other has a viscosity of \(2.987 \mathrm{~g} /(\mathrm{cm} \cdot \mathrm{s}) .\) Assign viscosities to 1 -pentanol and to hexane. Consider the compound \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\). Where do you think the boiling point of this compound might lie: above both 1 -pentanol and hexane, intermediate between these two compounds, or below both of these two compounds?

Briefly describe what it is that the Bragg equation relates?

Part 1: a Is it possible to add heat to a pure substance and not observe a temperature change? If so, provide examples. Describe, on a molecular level, what happens to the heat being added to a substance just before and during melting. Do any of these molecular changes cause a change in temperature? Part 2: Consider two pure substances with equal molar masses: substance A, having very strong intermolecular attractions, and substance \(\mathrm{B}\), having relatively weak intermolecular attractions. Draw two separate heating curves for 0.25 -mol samples of substance \(\mathrm{A}\) and substance \(\mathrm{B}\) in going from the solid to the vapor state. You decide on the freezing point and boiling point for each substance, keeping in mind the information provided in this problem. Here is some additional information for constructing the curves. In both cases, the rate at which you add heat is the same. Prior to heating, both substances are at \(-50^{\circ} \mathrm{C},\) which is below their freezing points. The heat capacities of \(\mathrm{A}\) and \(\mathrm{B}\) are very similar in all states. As you were heating substances \(\mathrm{A}\) and \(\mathrm{B}\), did they melt after equal quantities of heat were added to each substance? Explain how your heating curves support your answer. What were the boiling points you assigned to the substances? Are the boiling points the same? If not, explain how you decided to display them on your curves. c According to your heating curves, which substance reached the boiling point first? Justify your answer. Is the quantity of heat added to melt substance \(\mathrm{A}\) at its melting point the same as the quantity of heat required to convert all of substance \(\mathrm{A}\) to a gas at its boiling point? Should these quantities be equal? Explain.

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through the liquid at a given temperature and pressure. In an experiment, a 5.40-L sample of nitrogen gas, \(\mathrm{N}_{2}\), at \(20.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\) is bubbled through liquid isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O},\) at \(20.0^{\circ} \mathrm{C}\). Nitrogen containing the vapor of \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) at its vapor pressure leaves the vessel at \(20.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg} .\) It is found that \(0.6149 \mathrm{~g} \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) has evapo- rated. How many moles of \(\mathrm{N}_{2}\) are in the gas leaving the liquid? How many moles of alcohol are in this gaseous mixture? What is the mole fraction of alcohol vapor in the gaseous mixture? What is the partial pressure of the alcohol in the gaseous mixture? What is the vapor pressure of \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) at \(20.0^{\circ} \mathrm{C} ?\)

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