Chapter 6: Problem 61
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
Step by step solution
Understand the concept of specific heat
Convert mass to grams
Calculate the change in temperature
Calculate the energy change in joules
Calculate the final answer in kilojoules
Present the final calculation
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.
Energy Transfer Calculations
\[ Q = mc\Delta T \]
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
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
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.