/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 An ice cube with a mass of \(25 ... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 107

An ice cube with a mass of \(25 \mathrm{~g}\) at \(-18{ }^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds \(250 \mathrm{~mL}\) of hot water, initially at \(85^{\circ} \mathrm{C}\). What is the final temperature in the cup? The density of liquid water is \(1.00 \mathrm{~g} / \mathrm{mL}\); the specific heat capacity of ice is \(2.03 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C} ;\) the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{K} ;\) the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The final temperature in the cup is approximately \(75.18 掳C\).

Step by step solution

01

Calculate the energy to heat the ice from -18掳C to 0掳C

To calculate the energy required to heat the ice from -18掳C to 0掳C, we use the formula for energy transfer: \(q = mc\Delta T\) Where: \(q\) = energy transfer (J), \(m\) = mass of the ice (g), \(c\) = specific heat capacity of the ice (J/g掳C), and \(\Delta T\) = change in temperature (掳C). Here, \(m = 25 g\), \(c = 2.03 J/g掳C\), and \(\Delta T = 0掳C - (-18掳C) = 18 掳C\). Thus, the energy required to heat the ice to 0掳C is: \(q_1 = (25 g)(2.03 J/g 掳C)(18 掳C) = 913.5 J\)
02

Calculate the energy to melt the ice at 0掳C

To calculate the energy required to melt the ice at 0掳C, we use the enthalpy of fusion (\(Qf\)). Given that the enthalpy of fusion of water is \(6.01 kJ/mol\), we first need to determine the number of moles in the ice cube: Number of moles = \(\frac{mass}{molar\,mass}\) Since the molar mass of water is 18 g/mol, the number of moles in 25g of ice is: \(\frac{25 g}{18 g/mol} = 1.389 mol\) Now, we can find the energy required to melt the ice: \(q_2 = nQ_f = (1.389 mol)(6.01 kJ/mol) = 8.35 kJ = 8350 J\)
03

Set up the equation to solve for the final temperature

To solve for the final temperature, we equate the energy absorbed in heating the ice cube to the energy released by the hot water. The energy absorbed is the sum of energies (\(q_1\) and \(q_2\)) from steps 1 and 2. Let T be the final temperature, then the energy released by hot water is given by the formula: \(q_3 = m_{water}c_{water}\Delta T_{water}\) The mass of hot water can be determined from its volume and density: \(m_{water} = 250 \,mL * \frac{1.00\,g}{1\,mL} = 250\,g\) And, \(\Delta T_{water} = T_{initial} - T_{final} = 85 掳C - T\) Hence, the energy loss by the hot water can be represented as: \(q_3 = (250 g)(4.184 J/g 掳C)(85 掳C - T)\) Now we equate the energy absorbed in heating the ice cube (sum of \(q_1\) and \(q_2\)) to the energy released by the hot water (\(q_3\)): \(q_1 + q_2 = q_3\) \(913.5 J + 8350 J = (250 g)(4.184 J/g 掳C)(85 掳C - T)\) Now, solving the equation for the final temperature, T: \((85 掳C - T) = \frac{913.5 J + 8350 J}{(250 g)(4.184 J/g 掳C)}\) \((85 掳C - T) 鈮� 9.818 掳C\) Upon finding the difference: \(T 鈮� 85 掳C - 9.818 掳C 鈮� 75.18 掳C\) So, the final temperature in the cup is approximately \(75.18 掳C\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermochemistry that describes how much energy is needed to change the temperature of a substance. For ice, the specific heat capacity is given as \(2.03 \; \text{J/g}^\circ\text{C}\), which means it takes 2.03 Joules of energy to raise the temperature of one gram of ice by one degree Celsius. This property is crucial when calculating energy changes in substances, such as heating the ice from \(-18^\circ\text{C} \) to \(0^\circ\text{C}\).
We use the formula \( q = mc\Delta T \) to determine the energy transfer involved:
  • \( q \): energy (Joules)
  • \( m \): mass (grams)
  • \( c \): specific heat capacity
  • \( \Delta T \): temperature change (\circC)
Understanding specific heat capacity is key as it helps predict how substances will behave under temperature changes. It varies between materials; for example, water in liquid form has a specific heat capacity of \(4.184\; \text{J/g-K}\), allowing it to absorb heat efficiently, influencing everyday phenomena and technological applications.
Enthalpy of Fusion
The enthalpy of fusion is the amount of energy needed to turn a solid into a liquid at its melting point without changing its temperature. For water, this value is \(6.01 \text{kJ/mol}\). This indicates that it requires 6.01 kJ of energy to melt one mole of ice at \(0^\circ \text{C}\) into water.
In the exercise, after the ice is warmed to \(0^\circ \text{C}\), the next step is calculating how much energy is required to convert it to liquid water. This is done using the equation:
  • \( q_f = nQ_f \)
Where:
  • \( q_f \): the energy needed to melt the ice (Joules)
  • \( n \): number of moles of the substance
  • \( Q_f \): enthalpy of fusion (kJ/mol)
This energy calculation plays a crucial role in determining phase changes, enabling one to understand thermal dynamics in natural and industrial processes. It's a significant part of energy transfer calculations in thermochemical reactions.
Energy Transfer
Energy transfer is a critical concept in thermochemistry. It describes how energy moves from one system or object to another. In our scenario, energy is transferred from hot water to ice.
When the ice cube is added to the hot water, it absorbs energy and raises its temperature while simultaneously transferring energy to the ice, causing it to melt. How energy transfers can be calculated using the principle of conservation of energy. The energy lost by the hot water should equal the total energy the ice cube absorbs as it warms up and melts.
The relevant formula is:
  • \( q_\text{absorbed} = q_\text{released} \)
  • \( q_\text{absorbed} = q_1 + q_2 \) (heating + melting)
  • \( q_\text{released} = m c \Delta T \) for the hot water
Breaking down energy transfer into these steps makes it easier to solve complex problems and gain insights into energy dynamics in various reactions and processes, becoming a valuable tool in fields like chemistry, physics, and engineering.

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Most popular questions from this chapter

Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

The normal boiling point of the element mercury (Hg) is \(356.7{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol} .\) (a) When Hg boils at its nor- mal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).

Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)\)

Using data from Appendix \(\mathrm{C}\), calculate the change in Gibbs free energy for each of the following reactions. In each case, indicate whether the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AgCl}(s)\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{O}_{2}(g)\)
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