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 \((1 \mathrm{W}=1\) 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 \(\mathrm{cm} ?\)

Short answer

In a 1.00 square meter patch of ocean, approximately 3,213 g of water can be evaporated over a 12-hour day due to sunlight. The final temperature of the patch after being in sunlight for 12 hours, assuming no phase changes and sunlight penetration to a depth of 10.0 cm, is around 26.149 °C.

Step-by-step solution

Find the energy received by the surface in 12 hours

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

Calculate the grams of water evaporated due to sunlight

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.

Calculate the energy required to heat the water

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

Calculate the final temperature of the water

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.

Key concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding how substances respond to the addition of heat. It is defined as the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. For water, a commonly encountered substance, this value is fairly high, at 4.184 J/g°C.

This high specific heat capacity means water can absorb a lot of heat without a significant change in temperature. Consequently, water bodies, like oceans, can store and release large amounts of heat energy, influencing climate and weather patterns. When calculating temperature changes, it's important to multiply the specific heat capacity by the mass of the water and the temperature change to find the total energy absorbed or released.

Molar Mass

Molar mass is an essential aspect of chemistry, connecting fundamental units to real-world measurements. It represents the mass of one mole of a substance, expressed in grams per mole. For water, a critical substance in many calculations, the molar mass is 18.015 g/mol.

Knowing the molar mass allows for the conversion between moles and grams. This conversion is vital when calculating how much water evaporates due to energy from the sun. Since we deal with moles when using molecular or chemical equations, translating that into grams gives a tangible sense of quantity and helps bridge theoretical calculations with practical outcomes.

Energy Conversion

Energy conversion is pivotal when calculating heat transfer and understanding how energy impacts matter. The exercise involves transforming solar energy into the heat needed to evaporate water. Here, energy is given in Watts, where 1 Watt equals 1 Joule per second.

Over 12 hours, this energy accumulates and must be converted into Joules to be used in further calculations. This transformation helps determine how much water could evaporate with the input energy provided by the sunlight. Understanding energy conversion enables the effective mapping of energy flows and helps quantify physical changes in systems, especially in thermodynamics and environmental science.

Phase Changes

Phase changes represent fundamental physical transformations, such as water turning from liquid to vapor. During a phase change, a substance absorbs or releases energy without a change in temperature. In this exercise, the evaporation of water is the phase change of interest, driven by absorbed solar energy.

The energy needed for such a change is quantified by the enthalpy of evaporation. For water, this is 40.67 kJ/mol. Such transformations are critical in processes like weather dynamics, cooking, and industrial applications. Understanding phase changes, particularly how energy is used and conserved, helps explain diverse physical phenomena.

Final Temperature Calculation

Calculating the final temperature involves determining how much of the incoming energy not used for phase changes goes into increasing the water's temperature. Starting with the total energy from sunlight, one must subtract the energy used for evaporation.

The remaining energy heats the water, with its amount determined using the specific heat capacity formula. Remember to account for the mass of water and its specific heat to solve these calculations. Finally, add the temperature change to the initial temperature to find the new temperature after sunlight exposure. Such calculations are fundamental in thermodynamics, influencing everything from climate science to everyday heating systems.