Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 25.0 -mL sample of benzene at \(19.9^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C},\) and then frozen. How much heat was given off in this process? The density of benzene is \(0.80 \mathrm{g} / \mathrm{mL},\) its specific heat capacity is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) and its heat of fusion is \(127 \mathrm{J} / \mathrm{g}\).

Short Answer

Expert verified
The total heat given off is 3040.64 J.

Step by step solution

01

Calculate the Mass of Benzene

First, calculate the mass of benzene. Use the formula mass = volume × density. The given volume is 25.0 mL and the density is 0.80 g/mL. Thus, the mass of benzene is \( 25.0 \, \text{mL} \times 0.80 \, \text{g/mL} = 20.0 \, \text{g} \).
02

Calculate Heat Released in Cooling

Determine the amount of heat released when benzene is cooled from \(19.9^{\circ} \mathrm{C}\) to \(5.5^{\circ} \mathrm{C}\). Use the formula \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass (20.0 g), \( c \) is the specific heat capacity (1.74 J/g·K), and \( \Delta T \) is the change in temperature (\(5.5 - 19.9 = -14.4\, \text{K}\)). Calculate: \[ q = 20.0 \, \text{g} \times 1.74 \, \text{J/g⋅K} \times 14.4 \, \text{K} = 500.64 \, \text{J} \].
03

Calculate Heat Released in Freezing

Calculate the heat released during the phase change from liquid to solid at the melting point. Use the known heat of fusion. The formula is \( q = m \cdot \Delta H_{fus} \), where \( \Delta H_{fus} = 127 \, \text{J/g} \). Thus, \[ q = 20.0 \, \text{g} \times 127 \, \text{J/g} = 2540 \, \text{J} \].
04

Sum Total Heat Released

Add the heat released from cooling and the heat released from freezing to find the total heat released. \[ q_{total} = 500.64 \, \text{J} + 2540 \, \text{J} = 3040.64 \, \text{J} \].

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 Vaia!

Key Concepts

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

Specific Heat Capacity
The specific heat capacity is a critical concept in thermochemistry that helps us understand the amount of heat required to change the temperature of a substance. Specifically, it tells us how much energy is needed to raise the temperature of one gram of a substance by one Kelvin. In the exercise you encountered, benzene has a specific heat capacity of 1.74 J/g·K.

This means that for every 1 gram of benzene, it takes 1.74 joules of energy to increase its temperature by 1°C or 1 K (since the size of a degree on the Celsius scale is the same as that on the Kelvin scale).

Therefore, to calculate the heat energy released when benzene is cooled, we use the formula:
  • Energy (q) = mass (m) × specific heat capacity (c) × temperature change (ΔT)
In context, the specific heat capacity tells us how much energy will be absorbed or released during a temperature change. This is essential for calculating the heat changes in a system, like when benzene cools from 19.9°C to 5.5°C.
Heat of Fusion
Heat of fusion refers to the energy needed for a phase change from solid to liquid or vice versa, without changing the temperature. It's an important aspect of phase transitions in thermochemistry. In the case of benzene in the exercise, its heat of fusion is 127 J/g.

During a phase change at the melting point, benzene will require or release a specific amount of energy. Here, since we are interested in the freezing process, heat of fusion is the amount of heat released when 1 gram of benzene solidifies from liquid at its melting point, 5.5°C.

To calculate the total energy change during a phase transition for a given mass:
  • Energy (q) = mass (m) × heat of fusion (ΔHfus)
In our example, multiplying the mass of benzene (20.0 g) by its heat of fusion gives us the energy released during freezing.
Phase Change
A phase change is an occurrence when a substance transitions from one state of matter to another, such as from liquid to solid or vice versa. Unlike a simple temperature change, a phase change involves breaking or making the bonds that hold the particles together, thus requiring or releasing energy without a change in temperature.

In the exercise, the phase change of benzene occurs as it freezes at its melting point of 5.5°C. While benzene cools, it initially experiences a temperature change. Once it reaches its melting point, it undergoes the phase change from liquid to solid.

During this phase change, despite the energy exchange, the temperature remains constant until all the liquid has solidified. Understanding this principle is crucial in calculating the energy involved during the process using the heat of fusion.
Temperature Change Calculation
Calculating a temperature change involves finding the difference between the initial and final states of a substance. It is critical to understanding how much and in what way energy is absorbed or released in a thermochemical process.

In our case with benzene, initially at 19.9°C, and cooling down to 5.5°C, the temperature change (ΔT) is calculated by:
  • ΔT = final temperature - initial temperature
  • ΔT = 5.5°C - 19.9°C = -14.4°C
It's noteworthy that the change is negative, indicating a decrease in temperature. This decrease implies that energy is being released as benzene is cooled. Understanding and calculating ΔT precisely ensure accurate computation of heat changes using the formula q = m × c × ΔT.

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 add \(100.0 \mathrm{g}\) of water at \(60.0^{\circ} \mathrm{C}\) to \(100.0 \mathrm{g}\) of ice at \(0.00^{\circ} \mathrm{C} .\) Some of the ice melts and cools the water to \(0.00^{\circ} \mathrm{C} .\) When the ice and water mixture has come to a uniform temperature of \(0^{\circ} \mathrm{C},\) how much ice has melted?

The enthalpy change for the oxidation of styrene, \(\mathbf{C}_{8} \mathrm{H}_{8}\) is measured by calorimetry. $$\begin{aligned}\mathrm{C}_{8} \mathrm{H}_{8}(\ell)+10 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 8 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) & \\\& \Delta H_{\mathrm{rsn}}^{\circ}=-4395.0 \mathrm{kJ}\end{aligned}$$ Use this value, along with the standard heats of formation of \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell),\) to calculate the enthalpy of formation of styrene, in \(\mathrm{kJ} / \mathrm{mol}\).

Suppose that an inch of rain falls over a square mile of ground. (A density of \(1.0 \mathrm{g} / \mathrm{cm}^{3}\) is assumed.) The heat of vaporization of water at \(25^{\circ} \mathrm{C}\) is \(44.0 \mathrm{kJ} / \mathrm{mol} .\) Calculate the quantity of heat transferred to the surroundings from the condensation of water vapor in forming this quantity of liquid water. (The huge number tells you how much energy is "stored" in water vapor and why we think of storms as such great forces of energy in nature. It is interesting to compare this result with the energy given off, \(4.2 \times 10^{6} \mathrm{kJ},\) when a ton of dynamite explodes.)

The standard enthalpy of formation of solid barium oxide, \(\mathrm{BaO},\) is \(-553.5 \mathrm{kJ} / \mathrm{mol},\) and the enthalpy of formation of barium peroxide, \(\mathrm{BaO}_{2},\) is \(-634.3 \mathrm{kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic? $$\mathrm{BaO}_{2}(\mathrm{s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ (b) Draw an energy level diagram that shows the relationship between the enthalpy change of the decomposition of \(\mathrm{BaO}_{2}\) to \(\mathrm{BaO}\) and \(\mathrm{O}_{2}\) and the enthalpies of formation of \(\mathrm{BaO}(\mathrm{s})\) and \(\mathrm{BaO}_{2}(\mathrm{s})\).

You are on a diet that calls for eating no more than 1200 Cal/day. How many joules would this be?

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free