/*! 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 61 A vat of \(4.54 \mathrm{~kg}\) o... [FREE SOLUTION] | 91Ó°ÊÓ

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A vat of \(4.54 \mathrm{~kg}\) of water underwent a decrease in temperature from \(60.25^{\circ} \mathrm{C}\) to \(58.65^{\circ} \mathrm{C}\). How much energy in kilojoules left the water? (For this range of temperature, use a value of \(4.18 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) for the specific heat of water.

Short Answer

Expert verified
-30.256 kJ

Step by step solution

01

Understand the concept of specific heat

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. The formula to calculate the energy, Q, involved in a temperature change is given by: Q = mcΔT. where m = mass of substance, c = specific heat capacity, and ΔT = change in temperature.
02

Convert mass to grams

The mass of water is given in kilograms and needs to be converted to grams because the specific heat is given in terms of grams. Multiply the mass in kilograms by 1000 to get the mass in grams: 4.54 kg × 1000 g/kg = 4540 g.
03

Calculate the change in temperature

The change in temperature, ΔT, is the final temperature minus the initial temperature. Since the temperature decreased, ΔT = 58.65°C - 60.25°C = -1.60°C.
04

Calculate the energy change in joules

Substitute the mass in grams, the specific heat capacity, and the change in temperature into the formula to find the energy: Q = (4540 g) × (4.18 J/g°C) × (-1.60°C).
05

Calculate the final answer in kilojoules

The energy Q will come out in joules, but we want it in kilojoules. Since there are 1000 joules in a kilojoule, divide the result by 1000 to convert it: Q (in kilojoules) = Q (in joules) / 1000.
06

Present the final calculation

Perform the calculation from Steps 4 and 5: Q = 4540 g × 4.18 J/g°C × (-1.60°C) = -30256 J = -30.256 kJ (since 1 kJ = 1000 J).

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Key Concepts

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

Energy Transfer Calculations
Energy transfer in chemistry is a cornerstone concept that relates to the amount of heat exchanged with the surroundings during chemical processes. When substances undergo temperature changes, energy is either absorbed or released. This energy change can be quantified using the formula:

\[ Q = mc\Delta T \]
where \(Q\) represents the energy transfer in joules (J), \(m\) is the mass of the substance in grams (g), \(c\) is the specific heat capacity in joules per gram per degree Celsius (\(J/g^\circ C\)), and \(\Delta T\) is the change in temperature in degrees Celsius (\(^\circ C\)).

In a practical sense, this formula allows us to determine how much energy is required to heat up or cool down a substance. For instance, when heating a pot of water, energy from the stove is transferred to the water, resulting in a temperature increase. Conversely, when water in a container cools down, it loses energy to the environment, resulting in a temperature decrease. Precision in energy transfer calculations is crucial, as it can affect chemical reaction rates, phase changes, and various properties of materials.
Temperature Change in Chemistry
Temperature change in chemistry is often an indication of energy exchange between a system and its environment. It plays a critical role in understanding heat flow during chemical reactions and phase changes. The temperature change (\(\Delta T\)) is calculated by subtracting the initial temperature (\(T_{initial}\)) from the final temperature (\(T_{final}\)). It is essential to pay attention to the sign of \(\Delta T\) as it indicates whether the substance has gained or lost energy.

A positive \(\Delta T\) suggests that a substance's temperature rose, indicating energy was absorbed. A negative \(\Delta T\), as seen in our exercise, means energy was released. In laboratory settings, precise temperature measurements are crucial for calculating the energy changes accurately. Inconsistent measurements can lead to errors in the quantification of reaction enthalpies, activation energies, and in understanding the kinetics of reactions.
Heat Capacity of Water
The heat capacity of a substance is a measure of how much heat energy is required to raise its temperature by a given amount. Water has a high specific heat capacity, which means it can absorb or release a substantial amount of energy with only a small change in temperature—this characteristic is vital in a range of natural and industrial processes.

Water's specific heat capacity is approximately \(4.18 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\). This high value indicates water’s excellent ability to regulate temperature, which is why it is a good coolant and a stabilizing force for Earth's climate. In daily life, it's the reason why oceans and lakes temper the climate of coastal regions, and it's also why it takes a relatively long time for a pot of water to boil.

In the context of our original exercise, the knowledge of water’s specific heat capacity is crucial for calculating the energy that left the water as it cooled. This concept is also fundamental in environmental science and engineering, where the thermal properties of water affect ecosystem dynamics and energy management systems.

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Most popular questions from this chapter

How are heat and work defined?

A container filled with \(2.46 \mathrm{~kg}\) of water underwent a temperature change from \(25.24^{\circ} \mathrm{C}\) to \(27.31^{\circ} \mathrm{C}\). How much heat, measured in kilojoules, did the water absorb?

How do heat capacity and specific heat differ?

Calculate \(\Delta H^{\circ}\) in kilojoules for the following reaction, the preparation of the unstable acid nitrous acid, \(\mathrm{HNO}_{2}\). $$ \mathrm{HCl}(g)+\mathrm{NaNO}_{2}(s) \longrightarrow \mathrm{HNO}_{2}(l)+\mathrm{NaCl}(s) $$ Use the following thermochemical equations \(2 \mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HCl}(g)+\mathrm{Na}_{2} \mathrm{O}(s)\) \(\Delta H^{\circ}=+507.31 \mathrm{~kJ}\) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)\) $$ \begin{aligned} \Delta H^{\circ} &=-427.14 \mathrm{~kJ} \\ \mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) & \\ \Delta H^{\circ} &=-42.68 \mathrm{~kJ} \end{aligned} $$ \(2 \mathrm{HNO}_{2}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=+34.35 \mathrm{~kJ}\)

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.

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