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 \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .\) (a) Assuming that evaporation of water is only due 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 -hour day. (b) The specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). If the initial 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 hours, assuming no phase changes and assuming that sunlight penetrates uniformly to depth of \(10.0 \mathrm{~cm}\) ?

Step by step solution

01

Part (a): Calculate the energy received from sunlight

To calculate the energy received from sunlight, we'll use the formula: Energy = Power × Time We're given that the sunlight supplies 168 W per square meter and we're looking at a 1 square meter area for 12 hours. Time = 12 hours × 3600 seconds/hour = 43,200 seconds So the Energy from sunlight in 12 hours is: Energy = (168 J/s) × 43,200 s = 7,257,600 Joules

02

Part (a): Calculate the mass of water evaporated

To determine the mass of water evaporated, we will use the energy calculated above and the enthalpy of evaporation formula: Enthalpy of evaporation = (Energy / Mass of water) × Molar mass of water Given, the enthalpy of evaporation of water is 40.67 kJ/mol and molar mass of water is 18.015 g/mol. Rearrange the formula to calculate the mass of water, Mass of water = (Energy × Molar mass of water) / Enthalpy of evaporation Mass of water = (7,257,600 J × 18.015 g/mol) / (40.67 kJ/mol × 1000 J / kJ) Mass of water = 3,219.59 g Hence, 3,219.59 grams of water could be evaporated from a 1.00 square meter patch of ocean over a 12-hour day.

03

Part (b): Calculate the initial energy of the ocean patch

The initial energy can be calculated using the formula: Initial energy = Mass of water × Specific heat capacity × Initial temperature We are given that the initial temperature is 26°C and the specific heat capacity of water is 4.184 J/g°C. To calculate the mass of water in the patch, we need to use the volume and density of water: Density of liquid water = 1.0 g/cm³ Depth = 10.0 cm Area = 1.00 m² = 10000 cm² Volume of water in the patch = Area × Depth = 10000 cm² × 10.0 cm = 100,000 cm³ Mass of water in the patch = Volume × Density = 100,000 cm³ × 1.0 g/cm³ = 100,000 g Now, we can calculate the initial energy of the ocean patch: Initial energy = (100,000 g) × (4.184 J/g°C) × (26°C) = 10,879,200 J

04

Part (b): Calculate the energy of the ocean patch after 12 hours

The energy of the ocean patch after 12 hours of sunlight can be calculated by adding the energy supplied by sunlight: Energy after 12 hours = Initial energy + Energy from sunlight = 10,879,200 J + 7,257,600 J = 18,136,800 J

05

Part (b): Calculate the final temperature of the ocean patch

To calculate the final temperature of the ocean patch, we use the formula: Final temperature = Energy after 12 hours / (Mass of water × Specific heat capacity) Final temperature = (18,136,800 J) / (100,000 g × 4.184 J/g°C) Final temperature = 43.3°C Hence, the final temperature of the 1.00 square meter patch of ocean after being in sunlight for 12 hours and assuming no phase changes is 43.3°C.

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