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How many grams of coffee must evaporate from \(350 \mathrm{~g}\) of coffee in a \(100-\mathrm{g}\) glass cup to cool the coffee from \(95.0^{\circ} \mathrm{C}\) to \(45.0^{\circ} \mathrm{C}\) ? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is \(2340 \mathrm{~kJ} / \mathrm{kg}\) ( \(560 \mathrm{cal} / \mathrm{g}\) ). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.)

Short Answer

Expert verified
The mass of coffee that must evaporate is given by the expression: m_evap = (350 g * 4.184 J/g°C * 50.0°C) / 1,000 / 2.34 kJ/g.

Step by step solution

01

Calculate the heat lost by coffee while cooling

To find the amount of heat lost by the coffee as it cools down, use the specific heat capacity of water, which is approximately 4.184 Joules per gram per degree Celsius (J/g°C). The formula to calculate the heat (Q) lost is: Q = m * c * ΔT, where m is the mass of the coffee, c is the specific heat capacity, and ΔT is the change in temperature. Here, m = 350 g (mass of the coffee), c = 4.184 J/g°C, and ΔT = 95.0°C - 45.0°C = 50.0°C. Thus, Q = 350 g * 4.184 J/g°C * 50.0°C.
02

Convert the heat lost to kilojoules (kJ)

The heat lost calculated in step 1 will be in Joules. To convert it to kilojoules, which is the necessary unit to use the given heat of vaporization, divide by 1,000. If Q is the heat lost in Joules, then the heat lost in kilojoules is Q_kJ = Q / 1,000.
03

Calculate the mass of coffee that needs to evaporate

Using the heat of vaporization, calculate the mass of coffee (water) that must evaporate to lose the heat calculated in Step 2. The formula is: mass evaporated (m_evap) = Q_kJ / heat of vaporization. The heat of vaporization given is 2340 kJ/kg, which is equivalent to 2.34 kJ/g. Therefore, m_evap = Q_kJ / 2.34 kJ/g.
04

Calculate and report the final answer

Combine the results from previous steps to find the mass of coffee that needs to evaporate. Make sure the units are consistent and calculate the final mass in grams.

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Key Concepts

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

Specific Heat Capacity
Understanding the concept of specific heat capacity is crucial when analyzing how substances heat up or cool down. It's defined as the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. The specific heat capacity of water is approximately 4.184 Joules per gram per degree Celsius (\(4.184 \frac{J}{g \times ^{\text{o}}C}\)).

Water's high specific heat capacity means it can absorb a lot of heat before it begins to get hot. This property is why water is an effective coolant and used in many applications to regulate temperature.
In the exercise, we used this constant to calculate the heat lost by the coffee as it cooled. We multiplied the coffee's mass, the specific heat capacity of water, and the change in temperature to find this value. This calculation gives us insight into the amount of energy that would need to be removed in order for the coffee to reach the desired temperature.
Thermal Properties of Water
Water has unique thermal properties that make it stand out among other substances. Aside from its high specific heat capacity, water also has a high heat of vaporization, the amount of energy needed to convert one gram of a liquid into a gas at boiling point without changing its temperature.

These two properties are interrelated in our everyday experiences: think of how long it takes for a pot of water to boil or how sweat cools the body. The heat of vaporization for water is significantly higher than many other liquids, which is why it plays a critical role in Earth's climate and the regulation of temperature in living organisms.
In our exercise, knowing this property helped us understand the energy required for coffee (being treated as water for its similar thermal properties) to evaporate, which contributed to its cooling effect.
Energy Transfer in Thermodynamics
Thermodynamics is the study of energy and its transformations. One of its core principles is that energy can neither be created nor destroyed, but it can change forms, such as from kinetic to thermal energy, or transfer from one object to another.

In the context of our exercise, energy transfer occurred when the coffee, which started at a higher temperature, released heat to the surrounding environment as it cooled. This transfer of heat is governed by thermodynamic principles and is an example of energy changing form -- from the internal energy stored in the hot coffee to the energy carried away by the evaporated water molecules.
By analyzing this energy transfer, we could determine the mass of coffee that had to evaporate to achieve the desired cooling, illustrating the practical application of thermodynamic concepts in solving real-world problems.

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Most popular questions from this chapter

Show that the acceleration of any object down an incline where friction behaves simply (that is, where \(f_{\mathrm{k}}=\mu_{\mathrm{k}} N\) ) is \(a=g\left(\sin \theta-\mu_{\mathrm{k}} \cos \theta\right) .\) Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small \(\left(\mu_{\mathrm{k}}=0\right).\)

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