Question

The enthalpy of evaporation of water is \(40.67 \mathrm{~kJ} / \mathrm{mol}\). Sunlight striking Earth's surface supplies \(168 \mathrm{~W}\) per square meter \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .(\mathbf{a})\) Assuming that evaporation of water is due only to energy input from the Sun, calculate how many grams of water could be evaporated from a 1.00 square meter patch of ocean over a 12 -h day. (b) The specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). If the initial surface temperature of a 1.00 square meter patch of ocean is \(26^{\circ} \mathrm{C}\), what is its final temperature after being in sunlight for \(12 \mathrm{~h}\), assuming no phase changes and assuming that sunlight penetrates uniformly to depth of \(10.0 \mathrm{~cm} ?\)

Short answer

In summary, a 1.00 square meter patch of ocean can evaporate \(3212.07 \mathrm{~g}\) of water over a 12-hour day, considering only the energy input from the Sun. The final temperature of the ocean patch after being in sunlight for 12 hours would be \(27.73^{\circ} \mathrm{C}\), assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of \(10.0 \mathrm{~cm}\).

Step-by-step solution

Convert the energy received from sunlight into joules

First, we need to find out how much energy is received by our 1.00 square meter ocean patch during the 12-hour day. Since the energy input from sunlight is given in watts (W), we can multiply that by the total time of 12 hours (converted into seconds) to get the energy in joules. Keep in mind that 1 watt = 1 J/s. Energy received = 168 W/m² * 12 h * 3600 s/h = 7257600 J

Calculate the moles of water that can be evaporated

Now, we will find out how many moles of water can be evaporated using the energy provided by sunlight. To do this, we will divide the energy received (in joules) by the enthalpy of evaporation of water (in joules per mole). We must also convert the enthalpy of evaporation from kJ/mol to J/mol. Moles of water = Energy received / Enthalpy of evaporation = 7257600 J / (40.67 kJ/mol * 1000 J/kJ) = 178.31 mol

Calculate the grams of water evaporated

Now that we know how many moles of water can be evaporated, we can convert this value to grams using the molar mass of water (18.02 g/mol). Grams of water evaporated = Moles of water * Molar mass of water = 178.31 mol * 18.02 g/mol = 3212.07 g #Part 2: Calculating the final temperature of the ocean patch#

Calculate the mass of the ocean patch

To find the mass of the 1.00 square meter ocean patch, we need to use its depth (10.0 cm) and the density of water (1.0 g/cm³). Volume of water = 1.00 m² * 10.0 cm = 1000 cm³ Mass of water = Volume of water * Density of water = 1000 cm³ * 1.0 g/cm³ = 1000 g

Calculate the heat absorbed by the ocean patch

Using the energy received from sunlight, we will now calculate the heat absorbed by the ocean patch. Q = Energy received = 7257600 J = 7257.6 kJ

Calculate the change in temperature

Now, let's find out how much the temperature of the ocean patch increases as a result of absorbing the heat from sunlight. We can use the equation: Change in temperature = Q / (Mass of water * Specific heat capacity of water) = 7257.6 kJ / (1000 g * 4.184 J/g°C * 0.001 kJ/J) = 1.73°C

Calculate the final temperature of the ocean patch

Finally, we can determine the final temperature of the ocean patch by adding the initial temperature and the change in temperature. Final temperature = Initial temperature + Change in temperature = 26°C + 1.73°C = 27.73°C Thus, a 1.00 square meter ocean patch can evaporate 3212.07 g of water over a 12-hour day, and its final temperature after being in sunlight for 12 hours would be 27.73°C.

Key concepts

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates how much heat energy is needed to change the temperature of a given mass by 1°C. In the exercise, the specific heat capacity of liquid water is given as \(4.184 \, \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}\). This means that it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1°C.

When we calculate how sunlight affects the temperature of an ocean patch, the specific heat capacity helps us determine how much energy from the sun will change the water temperature. This is done using the formula:

• \[ \Delta T = \frac{Q}{m \times C} \]

Where \(\Delta T\) is the change in temperature, \(Q\) is the heat absorbed, \(m\) is the mass, and \(C\) is the specific heat capacity. For example, in this exercise, the change in temperature caused by the solar energy was calculated to be 1.73°C over the 12-hour sunlight period. This result reveals the efficiency of water's ability to absorb heat.

Molar Mass

Molar mass is a measure of the mass of a given substance (chemical element or compound) divided by the amount of substance. In the context of this problem, the molar mass of water is crucial for converting moles into grams. Water's molar mass is approximately 18.02 g/mol.

This property allows us to easily convert the moles of water, which we calculated based on the energy provided by the sun to evaporate it, into grams. Using the formula:

• Grams of water = Moles of water \(\times\) Molar mass of water

With the understanding that moles are a way to quantify the number of molecules, molar mass connects amount to mass, enabling us to calculate the 3212.07 grams of water evaporated from moles.

Energy Conversion

Energy conversion is a pivotal concept in thermodynamics and physical processes. In this exercise, sunlight provides energy to evaporate water from the ocean. Our starting point is the sun's power density of 168 watts per square meter.

To understand the full effect, we first convert this energy from watts into joules, because 1 watt is equal to 1 joule per second. For a continuous 12-hour exposure, the conversion is straightforward:

• Total Energy \( = 168 \, \mathrm{W/m}^2 \times 12 \, \mathrm{h} \times 3600 \, \mathrm{s/h} = 7257600 \, \mathrm{J} \)

This is a key step because joules are the standard units of energy for most calculations.

The next step in energy conversion uses the enthalpy of evaporation (40.67 kJ/mol converted to J/mol) to calculate the moles of water evaporated, thus showing how the incident solar energy translates directly into the phase change required to transform liquid water into vapor.

Phase Change

Phase change, specifically evaporation in this case, refers to the transition from liquid to gas. It requires energy to overcome the intermolecular forces holding the liquid together. The energy needed for this process is provided in the form of heat, which sunlight supplies.

The enthalpy of evaporation for water is 40.67 kJ/mol. This is the amount needed to convert 1 mole of water completely into vapor at a constant temperature. As the sunlight hits the water surface, it supplies heat that can change the water's state from liquid to vapor.

It's essential to understand that while some energy goes into increasing the temperature of the water (as per the specific heat capacity), this specific energy—the latent heat of evaporation—is that which directly causes phase change without affecting the temperature. This is why even on a sunny day, after water begins to boil, it stays at boiling point until all water has evaporated. In the exercise, the calculated energy converted enough water molecules (178.31 mol) from liquid to gaseous state, explaining why up to 3212.07 g of water was able to evaporate.