/*! 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 14 When 108 g of water at a tempera... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 14

When 108 g of water at a temperature of \(22.5^{\circ} \mathrm{C}\) is mixed with \(65.1 \mathrm{g}\) of water at an unknown temperature, the final temperature of the resulting mixture is \(47.9^{\circ} \mathrm{C}\). What was the initial temperature of the second sample of water?

Short Answer

Expert verified
The initial temperature of the second water sample was approximately \(89.9^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the Heat Transfer Equation

We use the concept of conservation of energy where the heat lost by the hotter water is equal to the heat gained by the cooler water. The formula to calculate heat is given by \( Q = m \, C \, \Delta T \), where \( Q \) is the heat absorbed or released, \( m \) is the mass, \( C \) is the specific heat capacity (which for water is \(4.18 \, \text{J/g} \, ^{\circ}\text{C}\)), and \( \Delta T \) is the change in temperature.
02

Apply the Heat Transfer Equation

For the 108g of water initially at \(22.5^{\circ} \mathrm{C}\), it gains heat and reaches \(47.9^{\circ} \mathrm{C}\). The change in temperature is \(47.9 - 22.5 = 25.4^{\circ} \mathrm{C}\). For the 65.1g of water, its temperature drops to \(47.9^{\circ} \mathrm{C}\) from an unknown initial temperature, \(T_2\). The change in temperature for this sample is \(T_2 - 47.9\).
03

Set Up the Heat Exchange Equation

The heat gained by the first sample is \( Q_1 = 108 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C} \times 25.4^{\circ} \mathrm{C} \). The heat lost by the second sample is \( Q_2 = 65.1 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C} \times (T_2 - 47.9)^{\circ} \mathrm{C} \). Set up the equation \(Q_1 = Q_2\) to solve for \(T_2\).
04

Solve the Equation for \(T_2\)

Calculate \( Q_1 = 108 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C} \times 25.4^{\circ} \mathrm{C} = 11445.76 \, \text{J} \). Substitute into the equation: \( 11445.76 = 65.1 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C} \times (T_2 - 47.9) \). Simplify this to solve for \(T_2\).
05

Calculate the Initial Temperature

Divide both sides by \(65.1 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C}\), which yields \(T_2 - 47.9 = \frac{11445.76}{272.418} \approx 42.03\). Adding \(47.9\) to both sides gives \(T_2 = 42.03 + 47.9 \approx 89.93^{\circ} \mathrm{C}\).
06

Verify and Conclude

Round \(T_2\) to one decimal place for precision, thus the initial temperature of the second sample of water was approximately \(89.9^{\circ} \mathrm{C}\). Confirm the steps and calculations to ensure accuracy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Conservation of Energy
The principle of conservation of energy is critical in understanding heat transfer. When two bodies at different temperatures come into contact, energy is transferred from the hotter body to the cooler one. This transfer continues until thermal equilibrium is reached, meaning both bodies are at the same temperature. In the context of our problem, energy is neither lost nor created; instead, it is exchanged between the two water samples. This is why the heat lost by the hotter water sample is equal to the heat gained by the cooler one.
Specific Heat Capacity
Specific heat capacity is a measure of how much energy is needed to raise the temperature of a given mass of a substance by one degree Celsius. For water, this value is relatively high at 4.18 J/g掳C, which means it takes 4.18 joules of energy to increase the temperature of one gram of water by one degree Celsius. In our problem, knowing the specific heat capacity of water allows us to calculate the amount of heat transferred when the water samples mix, using the formula: \(Q = m \cdot C \cdot \Delta T\). This formula essentially states that the heat transferred \(Q\) is the product of mass \(m\), specific heat capacity \(C\), and the change in temperature \(\Delta T\).
Temperature Change
Temperature change is a key component of the heat transfer equation. It is calculated as the difference between the initial and final temperatures of a substance. In our exercise, the 108 g sample of water experiences a temperature change from \(22.5^{\circ} \mathrm{C}\) to \(47.9^{\circ} \mathrm{C}\), a positive change since it gained heat. Meanwhile, the temperature change for the 65.1 g sample is calculated relative to the unknown initial temperature \(T_2\), and final temperature \(47.9^{\circ} \mathrm{C}\). Understanding temperature changes helps determine how much energy has been transferred during the process.
Heat Exchange Equation
The heat exchange equation is fundamental in solving problems like ours. It is represented by the formula \(Q_1 = Q_2\), where \(Q_1\) is the heat gained by one sample, and \(Q_2\) is the heat lost by the other. By substituting the known values into this equation, we can solve for any unknowns. In this case, the unknown is the initial temperature \(T_2\) of the 65.1 g water sample. By equating the heat gained and lost (since energy transfer is conserved), and using the given mass, specific heat capacity, and calculated temperature change, we form an equation that we can solve for \(T_2\). This step-by-step approach leads us to accurately determine that \(T_2\) is approximately \(89.9^{\circ} \mathrm{C}\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You drink \(350 \mathrm{mL}\) of diet soda that is at a temperature of \(5^{\circ} \mathrm{C}\) (a) How much energy will your body expend to raise the temperature of this liquid to body temperature \(\left(37^{\circ} \mathrm{C}\right) ?\) Assume that the density and specific heat capacity of diet soda are the same as for water. (b) Compare the value in part (a) with the caloric content of the beverage. (The label says that it has a caloric content of 1 Calorie.) What is the net energy change in your body resulting from drinking this beverage? (1 Calorie = \(1000 \mathrm{kcal}=4184 \mathrm{J} .)\) (c) Carry out a comparison similar to that in part (b) for a nondiet beverage whose label indicates a caloric content of 240 Calories.

For each of the following, define a system and its surroundings, and give the direction of energy transfer between system and surroundings. (a) Methane burns in a gas furnace in your home. (b) Water drops, sitting on your skin after a swim, evaporate. (c) Water, at \(25^{\circ} \mathrm{C},\) is placed in the freezing compartment of a refrigerator, where it cools and eventually solidifies. (d) Aluminum and \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) are mixed in a flask sitting on a laboratory bench. A reaction occurs, and a large quantity of energy is evolved as heat.

These questions are not designated as to type or location in the chapter. They may combine several concepts. The following terms are used extensively in thermodynamics. Define each and give an example. (a) exothermic and endothermic (b) system and surroundings (c) specific heat capacity (d) state function (e) standard state (f) enthalpy change, \(\Delta H\) (g) standard enthalpy of formation

Enthalpy Changes Nitrogen monoxide, a gas recently found to be involved in a wide range of biological processes, reacts with oxygen to give brown \(\mathrm{NO}_{2}\) gas. $$ \begin{aligned} 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow & 2 \mathrm{NO}_{2}(\mathrm{g}) \\ \Delta_{r} H^{\circ} &=-114.1 \mathrm{kJ} / \mathrm{mol}-\mathrm{pxn} \end{aligned} $$ Is this reaction endothermic or exothermic? What is the enthalpy change if \(1.25 \mathrm{g}\) of \(\mathrm{NO}\) is converted completely to \(\mathrm{NO}_{2} ?\)

A \(0.692-\mathrm{g}\) sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) was burned in a constant-volume calorimeter. The temperature rose from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C}\). The calorimeter contained \(575 \mathrm{g}\) of water, and the bomb had a heat capacity of \(650 \mathrm{J} / \mathrm{K}\). What is \(\Delta U\) per mole of glucose?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.