Question

The enthalpy of evaporation of water is $40.67\mathrm{kJ}/\mathrm{mol}$. Sunlight striking Earthis surface supplies 168 W per square meter $(1\mathrm{W}=1\mathrm{watt}=1\mathrm{J}/\mathrm{s})$. (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

(a) Approximately 3,214.9 grams of water could be evaporated from a $1.00$ square meter patch of ocean over a 12-hour day. (b) The final temperature of the water after being in sunlight for 12 hours is ${43.34}^{\circ }\text{C}$.

Step-by-step solution

Calculate the received energy per unit of volume of water

Over a 12-hour day, the total energy received by a $1.00m^2$ patch of ocean is given by: $$E=P\cdot A\cdot t$$ where $E$ is the energy, $P$ is the power supplied by sunlight per square meter (168 W), $A$ is the area (1 m${}^2$), and $t$ is the time (12 h). First, we need to convert the time from hours to seconds: $$12\text{hours}\times \frac{3600\text{s}}{1\text{hour}}=43200\text{s}$$ Now, we can calculate the total energy received: $$E=(168\text{W})(1{\text{m}}^2)(43200\text{s})=7.2576\times {10}^6\text{J}$$

Calculate the mass of water evaporated

To find the mass of water evaporated, we'll use the formula: $$m=\frac{E}{H_v}$$ where $m$ is the mass of water evaporated, $E$ is the total energy received, and $H_v$ is the enthalpy of evaporation per mol of water (40.67 kJ/mol). First, let's convert the enthalpy of evaporation to J/mol: $$40.67\text{kJ/mol}\times \frac{{10}^3\text{J}}{1\text{kJ}}=40,670\text{J/mol}$$ Now, we can calculate the mass of water evaporated in moles: $$m_{\text{mol}}=\frac{7.2576\times {10}^6\text{J}}{40,670\text{J/mol}}=178.5\text{mol}$$ Finally, convert the mass in moles to grams using the molar mass of water (18.015 g/mol): $$m_{\text{g}}=(178.5\text{mol})(18.015\text{g/mol})=3,214.9\text{g}$$ So, about 3,214.9 grams of water could be evaporated from a $1.00$ square meter patch of ocean over a 12-hour day.

Calculate the final temperature of the water

To find the final temperature after being in sunlight for 12 hours, we'll use the formula: $$Q=mc\mathrm{\Delta }T$$ where $Q$ is the absorbed energy, $m$ is the mass of water, $c$ is the specific heat capacity of water, and $\mathrm{\Delta }T$ is the temperature change. First, we need to find the mass of water in the $1.00$ square meter patch with $10cm$ depth. The volume of water is: $$V=A\cdot d=(1{\text{m}}^2)(0.1\text{m})=0.1{\text{m}}^3$$ Assuming the density of water as $\rho =1000{\text{kg/m}}^3$, $$m_{\text{water}}=V\cdot \rho =(0.1{\text{m}}^3)(1000{\text{kg/m}}^3)=100\text{kg}$$ Next, convert the mass of water to grams: $$m_{\text{water\_g}}=(100\text{kg})\times \frac{1000\text{g}}{1\text{kg}}=100,000\text{g}$$ Now, using the specific heat capacity of water, $c=4.184{\text{J/g}}^{\circ }\text{C}$, and the received energy, $E=7.2576\times {10}^6\text{J}$, $$\mathrm{\Delta }T=\frac{Q}{mc}=\frac{7.2576\times {10}^6\text{J}}{(100,000\text{g})(4.184{\text{J/g}}^{\circ }\text{C})}={17.34}^{\circ }\text{C}$$ Now, just add the initial temperature, $T_{\text{initial}}={26}^{\circ }\text{C}$, to calculate the final temperature: $$T_{\text{final}}=T_{\text{initial}}+\mathrm{\Delta }T={26}^{\circ }\text{C}+{17.34}^{\circ }\text{C}={43.34}^{\circ }\text{C}$$ In conclusion, (a) approximately 3,214.9 grams of water could be evaporated from a $1.00$ square meter patch of ocean over a 12-hour day, and (b) the final temperature of the water after being in sunlight for 12 hours is ${43.34}^{\circ }\text{C}$.

Key concepts

Specific Heat Capacity

Specific heat capacity is a property of a substance that describes how much energy is required to raise the temperature of one gram of the substance by one degree Celsius. For water, this value is quite high at 4.184 J/g°C. This means water can absorb a lot of heat without a significant rise in temperature, which is why it's effective at regulating temperatures in the environment.

When dealing with the specific heat capacity, we use the formula for heat transfer:

• $Q=mc\mathrm{\Delta }T$

Here, $Q$ represents the heat absorbed or released, $m$ is the mass, $c$ is the specific heat capacity, and $\mathrm{\Delta }T$ is the change in temperature. This formula helps us understand how much the temperature of a substance will change when subjected to a certain amount of energy.
By applying this to the ocean’s surface, we calculated that after absorbing sunlight energy for 12 hours, the water's temperature would rise, demonstrating how specific heat capacity affects environmental and climatic processes.

Energy Conversion

Energy conversion involves transforming energy from one form to another. In this case, sunlight energy (solar energy) is converted into thermal energy absorbed by the ocean. We measure the amount of sunlight energy the ocean receives with the power formula:

• $E=P\cdot A\cdot t$

where $E$ is the energy received, $P$ is the power (in watts), $A$ is the area, and $t$ is the time the sunlight continues to shine. Power is energy per time unit, and since 1 Watt equals 1 Joule per second, we convert hours into seconds for our calculations to match units.

After finding how much total energy the ocean patch receives, we explored how this energy could heat the water and potentially lead to evaporation. As energy is absorbed, it changes the current state of the substance, converting the energy into either a temperature increase or a phase change.

Evaporation Calculation

Evaporation is when a liquid turns into a gas due to added energy, often from heat. The enthalpy of evaporation is a key concept here, representing the needed energy to evaporate a mole of substance, in this case, 40.67 kJ/mol for water.

• Evaporation uses the energy absorbed by water to overcome molecular bonds.

To calculate how much water evaporates, we first determine how much energy the water has absorbed. Using the formula:

• $m=\frac{E}{H_v}$

where $m$ is the mass of water evaporated, $E$ is the energy received, and $H_v$ is the enthalpy of evaporation. In this problem, after receiving $7.2576\times {10}^6$ J of energy, we calculated how the energy translates to the mass of evaporated water.

Understanding evaporation calculations is crucial as it explains phase changes in nature, affecting weather patterns and climate systems by showing how energy transforms the physical states of matter.