Strong sunshine bombards the Earth with about \(1 \mathrm{~kJ} \cdot \mathrm{m}^{-2}\) in \(1 \mathrm{~s}\). Calculate the maximum mass of pure ethanol that can be vaporized in 10 min from a beaker left in strong sunshine, assuming the surface area of the ethanol to be \(50 \mathrm{~cm}^{2}\). Assume all the heat is used for vaporization, not to increase the temperature.

Short Answer

Expert verified
The maximum mass of ethanol that can be vaporized is the total energy absorbed divided by the latent heat of vaporization of ethanol.

Step by step solution

01

Convert the sunshine energy to joules per 10 minutes

First, convert the surface area from square centimeters to square meters by dividing by 10,000 (since there are 100 cm in a meter and therefore 10,000 square centimeters in a square meter). Then, calculate the total energy absorbed by the ethanol over 10 minutes (600 seconds), by multiplying the provided energy flux by the surface area in square meters and the time in seconds.
02

Calculate the heat required to vaporize ethanol

The heat required to vaporize a substance is given by the formula Q = m * Lv,where Q is the heat supplied, m is the mass of the substance, and Lv is the latent heat of vaporization. Use the latent heat of vaporization for ethanol, which is approximately 841 kJ/kg.
03

Calculate the maximum mass of ethanol vaporized

Rearrange the formula to solve for m:m = Q / Lv.Now, substitute the total energy absorbed by the ethanol (calculated in Step 1) for Q, and the latent heat of vaporization of ethanol for Lv, to find the maximum mass of ethanol that can be vaporized.

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Key Concepts

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

Latent Heat of Vaporization
The concept of latent heat of vaporization is crucial in understanding physical changes of states, especially when a liquid turns into a gas. This property is a measure of the amount of heat energy required to change one kilogram of a liquid into a gas at constant temperature. Essentially, it is the heat needed to overcome the molecular forces that hold the liquid together, without changing its temperature.

For substances like ethanol, the latent heat of vaporization has a specific value. This value is relatively high because the forces between ethanol molecules require significant energy to overcome. This is why knowing the latent heat of vaporization is essential when calculating how much energy is needed to vaporize a given mass of ethanol, or any liquid for that matter. In the provided exercise, to find out the mass of ethanol that can be vaporized, we must use this constant along with the amount of energy available from the sun.
Energy Conversion
In the problem at hand, energy conversion plays a pivotal role in finding the solution. Energy conversion refers to the process of changing energy from one form to another. In our scenario, solar energy is being converted into thermal energy, which then is used to vaporize the ethanol.

The conversion starts when the sunlight, which is radiant energy, strikes the surface of the ethanol. The radiant energy is absorbed by the ethanol and becomes thermal energy that specifically goes toward vaporizing the ethanol, rather than raising its temperature. The exercise assumes that all the solar energy is converted into the energy required for vaporization, which simplifies the calculation. In real-world scenarios, some energy might be lost or used for a different purpose, but this assumption keeps our calculation straightforward.
Heat Transfer
Heat transfer is a fundamental concept that describes how thermal energy moves from one place to another. It can occur in three different ways: conduction, convection, and radiation. In our context, the most relevant type of heat transfer is radiation, specifically from the sun to the surface of the ethanol.

Understanding this concept is vital for the exercise because it helps us grasp how the given energy from the sun is being used to vaporize the ethanol. The energy from the sun irradiates onto the ethanol surface, transferring heat to the liquid. As we focus on this transfer process, we are interested in the amount of energy transferred over a certain area, in this case, 50 cm squared, and over a specific timespan — 10 minutes. By making careful calculations based on the principles of heat transfer, we can predict how much ethanol will vaporize under these conditions.
Chemical Thermodynamics
Chemical thermodynamics deals with the relationship between heat (or energy) and work within chemical processes. It's a broad field that includes concepts like the first law of thermodynamics, which in essence states that energy cannot be created or destroyed, only transferred or transformed. This principle is implicitly applied in our exercise, as the solar energy is neither being created nor destroyed but rather converted and transferred into the process of vaporizing the ethanol.

When solving problems related to vaporization, it's essential to remember these principles because they help us ensure energy conservation is accounted for in our calculations. In the context of the exercise, by using chemical thermodynamics concepts, we can demonstrably show that the energy coming from the sun is the driving force for vaporizing the ethanol, subtly underscoring the law of conservation of energy in a practical application.

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Most popular questions from this chapter

(a) At its boiling point, the vaporization of \(0.235 \mathrm{~mol} \mathrm{CH}_{4}(1)\) requires \(1.93 \mathrm{~kJ}\) of heat. What is the enthalpy of vaporization of methane? (b) An electric heater was immersed in a flask of boiling crhanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), and \(22.45 \mathrm{~g}\) of ethanol was vaporized when \(21.2 \mathrm{~kJ}\) of energy was supplied. What is the enthalpy of vaporization of ethanol?

Calculate the lattice enthalpy of solid potassium bromide, \(\mathrm{KBr}(\mathrm{s}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{Br}^{-}(\mathrm{g})\), from the following information: $$ \begin{aligned} &\Delta H_{i}{ }^{\circ}\left(\mathrm{KBr}, \text { s) }=-394 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right. \\ &\Delta H_{i}{ }^{\circ}(\mathrm{K}, \mathrm{g})=+89.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ First ionization energy of \(\mathrm{K}(\mathrm{g})=+425.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta H_{\text {vap }}{ }^{\circ}\left(\mathrm{Br}_{2}, \mathrm{l}\right)=+30.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Br-Br bond dissociation cuthalpy \(=+192.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Electron attachment to \(\mathrm{Br}(\mathrm{g})\) : $$ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{Be}^{-}(\mathrm{g}), \quad \Delta H^{*}=-331.0 \mathrm{~kJ} $$

Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride, \(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{~s})\), from the following data: $$ \begin{array}{ll} 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \\ & \Delta H^{\circ}=-1049 \mathrm{~kJ} \\ \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{aq}) & \Delta H^{\circ}=-74.8 \mathrm{~kJ} \\ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g}) & \Delta H^{\circ}=-185 \mathrm{~kJ} \\ \mathrm{AlCl}_{3}(\mathrm{~s}) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{aq}) & \Delta H^{\circ}=-323 \mathrm{~kJ} \end{array} $$

Write the thermochemical equations that give the values of the standard enthalpies of formation for (a) \(\mathrm{KClO}_{3}\) (s), potassium chlorate; (b) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH}\) (s), glycine(s); (c) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})\), alumina.

Near room temperature, the specific heat capacity of benzene is \(1.05 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). Calculate the heat nceded to raise the temperature of \(50.0 \mathrm{~g}\) of benzene from \(25.3^{\circ} \mathrm{C}\) to \(37.2^{\circ} \mathrm{C}\). (b) A \(1.0-\mathrm{kg}\) block of aluminum is supplied with \(490 \mathrm{~kJ}\) of heat. What is the temperature change of the aluminum? The specific heat capacity of aluminum is \(0.90 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \cdot \mathrm{g}^{-1}\).

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