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 \((1 \mathrm{W}=1\) watt \(=1 \mathrm{J} / \mathrm{s})\) . (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 \(\mathrm{cm} ?\)

First, we need to find out how much energy the 1.00 square meter patch of ocean receives in a 12-hour period from the Sun. We are given a power input of 168 W per square meter. To find the energy, we will simply multiply the power input by the time in seconds: Energy received = Power input × Time Time = 12 hours = 12 × 60 × 60 seconds Now we calculate the energy received from sunlight: Energy received = 168 J/s × 12 × 60 × 60 s = 7,257,600 J

Now we will convert this energy received into the grams of water evaporated. For this, we can use the enthalpy of evaporation given as 40.67 kJ/mol. We can first convert this energy into Joules: Enthalpy of evaporation = 40.67 kJ/mol × 1,000 J/kJ = 40,670 J/mol Now, divide the energy received by the enthalpy of evaporation to find the moles of water evaporated: Moles of water evaporated = 7,257,600 J / 40,670 J/mol = 178.37 mol Next, convert moles of water into grams using the molar mass of water, which is 18.015 g/mol: Grams of water evaporated = 178.37 mol × 18.015 g/mol ≈ 3,213 g So, 3,213 g of water will be evaporated from the 1.00 square meter patch of ocean over a 12-hour day due to sunlight.

Since the sunlight penetrates to a depth of 10.0 cm (0.1 m), first, we need to find the volume of water that is being heated: Volume = Area × Depth = 1.00 m² × 0.1 m = 0.1 m³ Now, convert this volume to mass, using the density of water (1,000 kg/m³): Mass of water = Volume × Density = 0.1 m³ × 1,000 kg/m³ = 100 kg = 100,000 g

Since the energy input is limited to the available sunlight, and we have already calculated the grams of water evaporated, we can now calculate the remaining energy available to heat the water. We first need to convert the grams of water evaporated back into energy: Energy used to evaporate water = Grams of water evaporated × (Enthalpy of evaporation / Molar mass) = 3,213 g × (40,670 J/mol / 18.015 g/mol) ≈ 7,195,274 J Now, subtract the energy used for evaporation from the total energy received to find the energy available to heat the remaining water: Remaining energy = Energy received - Energy used to evaporate water = 7,257,600 J - 7,195,274 J ≈ 62,326 J Finally, use the equation for heating a substance with specific heat capacity to find the change in temperature: ΔT = Q / (mass × specific heat capacity) ΔT ≈ 62,326 J / (100,000 g × 4.184 J/g°C) ≈ 0.149 °C Now, add the change in temperature to the initial temperature to find the final temperature: Final temperature = Initial temperature + ΔT = 26 °C + 0.149 °C ≈ 26.149 °C The final temperature of the 1.00 square meter patch of ocean after being in sunlight for 12 hours, assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of 10.0 cm, is approximately 26.149 °C.