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 \(10.0 \mathrm{~cm} ?\)

Step by step solution

01

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

02

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

03

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#

04

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

05

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

06

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

07

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.

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