Problem 39 - Chemistry The Central Science In Si Units

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 \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .\) (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 \(12-\mathrm{h}\) 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

Answer

(a) Approximately 3215 grams of water could be evaporated. (b) The final temperature would be 43.35°C.

Step 1: Calculate Total Energy Input from Sunlight

First, calculate the total energy input from the sunlight over 12 hours for a 1 square meter area. The power supplied by sunlight is 168 W per square meter, and 1 W is equivalent to 1 J/s. Therefore, \[\text{Energy (J)} = 168 \, \text{W/m}^2 \times (12 \times 60 \times 60) \, \text{s} = 168 \, \text{W/m}^2 \times 43200 \, \text{s} = 7,257,600 \, \text{J/m}^2\]

Step 2: Convert Energy to Moles of Water Evaporated

The energy required to evaporate one mole of water is 40.67 kJ/mol, which is 40670 J/mol. To find out how many moles of water can be evaporated:\[\text{Moles of water} = \frac{7,257,600 \, \text{J}}{40670 \, \text{J/mol}} \approx 178.45 \, \text{mol}\]

Step 3: Convert Moles to Grams

Water has a molar mass of approximately 18.015 g/mol. To convert moles to grams:\[\text{Mass of water (g)} = 178.45 \, \text{mol} \times 18.015 \, \text{g/mol} \approx 3215.02 \, \text{g}\]

Step 4: Calculate Volume of Water Heated

The sunlight penetrates uniformly to 10.0 cm, or 0.1 m in depth. The volume of this water which is 1 square meter wide is:\[\text{Volume} = 1.00 \, \text{m}^2 \times 0.1 \, \text{m} = 0.1 \, \text{m}^3 = 100 \, \text{L} \times 1000 \, \text{g/L} = 100,000 \, \text{g}\]

Step 5: Calculate Temperature Change Using Specific Heat Capacity

Using the specific heat capacity, find the change in temperature. The formula for heat is:\[q = m \cdot c \cdot \Delta T\]Rearranging for \(\Delta T\):\[\Delta T = \frac{q}{m \cdot c} = \frac{7,257,600 \, \text{J}}{100,000 \, \text{g} \times 4.184 \, \text{J/g}^{\circ}\mathrm{C}} \approx 17.35^{\circ}\mathrm{C}\]

Step 6: Calculate Final Temperature

Add the change in temperature to the initial temperature to get the final temperature:\[\text{Final Temperature} = 26^{\circ} \text{C} + 17.35^{\circ} \text{C} = 43.35^{\circ} \text{C}\]

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of heat required to change the temperature of a unit mass of the substance by one degree Celsius. For water, this value is quite significant at 4.184 J/g°C. This means, when heating a gram of water by 1°C, you need to supply 4.184 Joules of energy.

This concept helps us understand how water in oceans or lakes reacts to temperature changes throughout the day.

• Explains why water has a moderating effect on climate.
• Determines how quickly water bodies can heat up or cool down.

When calculating changes in temperature, the specific heat capacity allows us to quantify how much the temperature of water will change for a given amount of energy.

Energy Input

Energy input, in this context, refers to the energy received from the Sun by Earth's surface, specifically through sunlight. This exercise uses a fixed power value of 168 W/m², which is the energy received per second over each square meter.

Given a full 12-hour day (or 43,200 seconds), we calculate the total energy input.
This is crucial:

• To gauge the potential effects of sunlight on surface temperatures.
• To determine how much water could evaporate from natural bodies like oceans.

This energy supply is vital for calculations involving changes in temperature and phase changes like evaporation.

Evaporation Calculation

Calculating evaporation involves understanding how sunlight can cause water to transition from a liquid phase to a gas. This requires a specific amount of energy, known as the enthalpy of evaporation, which for water is 40.67 kJ/mol.

By dividing the total energy input by this enthalpy, we determine the number of moles of water that can evaporate. Once we have moles, converting these to grams by using water's molar mass (18.015 g/mol) gives the volume of water evaporated. This approach outlines the detailed quantification of water loss due to solar energy input.

Temperature Change

The temperature change of a body of water due to solar energy is determined by the heat absorbed and the specific heat capacity of water. The formula, represented as \( q = m \cdot c \cdot \Delta T \), relates heat energy \( q \) to mass \( m \), specific heat capacity \( c \), and temperature change \( \Delta T \).

In the exercise example, this formula shows how exposure to sunlight changes the water's temperature. Given the energy supply and water properties, you can calculate the temperature increase over a defined period.

• Helps us understand natural temperature fluctuations in oceans or pools.
• Determines the heating effects experienced during long sun exposure.
• Aids in understanding the balance between energy absorption and temperature rise.