Problem 84 - Fundamentals of Physics

Question

A \(400 \mathrm{~W}\) immersion heater is placed in a pot containing \(2.00 \mathrm{~L}\) of water at \(20^{\circ} \mathrm{C}\). (a) How long will the water take to rise to the boiling temperature, assuming that \(80 \%\) of the available energy is absorbed by the water? (b) How much longer is required to evaporate half of the water?

Step-by-Step Solution

Answer

(a) 1339.52 s to boil; (b) additional 4520 s to evaporate half the water.

Step 1: Calculate the Energy Needed to Heat the Water

First, compute the energy needed to raise the temperature of the water from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). Use the formula for energy: \[ Q_1 = m \cdot c \cdot \Delta T \]where \( m \) is the mass of water, \( c \) is the specific heat capacity of water \((4.186 \mathrm{~J/g^{\circ}C})\), and \( \Delta T \) is the temperature change \((100^{\circ} C - 20^{\circ} C = 80^{\circ} C )\). The mass of water \(m\) is \(2.00\mathrm{~kg} = 2000\mathrm{~g}\), for 2 liters of water. Therefore, \[ Q_1 = 2000 \mathrm{g} \times 4.186 \mathrm{~J/g^{\circ}C} \times 80^{\circ} C = 669,760 \mathrm{~J} \].

Step 2: Adjust Energy for Efficiency

The actual energy absorbed by the water is only 80% of the total energy input. Thus,\[ Q_{effective} = 0.8 \times Q_1 = 0.8 \times 669,760 \mathrm{~J} = 535,808 \mathrm{~J} \].

Step 3: Calculate the Time to Boil

The power of the heater is 400 W, which is equivalent to 400 J/s. The time \( t \) taken is given by \[ t = \frac{Q_{effective}}{P} \],where \( Q_{effective} \) is the energy absorbed and \( P \) is the power. Thus,\[ t = \frac{535,808 \mathrm{~J}}{400 \mathrm{~J/s}} = 1339.52 \mathrm{~s} \].

Step 4: Calculate the Energy Needed to Evaporate Half of the Water

To find the energy needed to evaporate half of the 2.00 L (1.00 L of water), we need to calculate the heat needed for phase change. The latent heat of vaporization for water is \(2260 \mathrm{~J/g}\). The mass of 1.00 L water is \(1000\mathrm{~g}\). Thus, \[ Q_2 = m \cdot L = 1000 \mathrm{~g} \times 2260 \mathrm{~J/g} = 2,260,000 \mathrm{~J} \].

Step 5: Adjust Energy for Efficiency

Again, considering the efficiency of 80%, the energy absorbed for evaporation is \[ Q_{effective} = 0.8 \times 2,260,000 \mathrm{~J} = 1,808,000 \mathrm{~J} \].

Step 6: Calculate the Additional Time to Evaporate Half of the Water

Determine the additional time \( t \) required using the heater:\[ t = \frac{Q_{effective}}{P} = \frac{1,808,000 \mathrm{~J}}{400 \mathrm{~J/s}} = 4520 \mathrm{~s} \].

Key Concepts

Specific Heat Capacity

Specific heat capacity is an important concept when it comes to understanding how substances absorb heat. Basically, it tells us how much energy we need to raise the temperature of a specific amount of a substance by one degree Celsius. It is a measure of a material's ability to hold thermal energy compared to its mass. In the exercise, specific heat capacity helps us understand how much energy, in joules, the water requires to increase its temperature from room temperature to boiling point.

Water has a high specific heat capacity, which is why it takes a lot of energy to change its temperature. Specifically, the specific heat capacity of water is 4.186 J/g°C. This means you would need 4.186 joules of energy to raise 1 gram of water by 1 degree Celsius. Because of this, water can absorb or release a great deal of heat without experiencing a large temperature change.

• It explains why oceans and lakes can store and retain heat.
• It contributes to the ability of water to regulate climate and weather.

Latent Heat of Vaporization

The latent heat of vaporization is the amount of heat required to change a unit mass of a liquid into vapor at constant temperature. This concept focuses on the energy needed for a phase change, not a temperature change. In this exercise, it addresses the heat required to convert half of the water in the pot from liquid to vapor, hence increasing its energy without altering its temperature.

For water, the latent heat of vaporization is 2260 J/g. This means to convert 1 gram of water at its boiling point into steam without increasing its temperature, it takes 2260 joules of energy.

• It can help explain phenomena like sweating, where the body's heat is used up in converting sweat into vapor, cooling us down.
• This concept is imperative in processes such as distillation and refrigeration.

Power and Energy Efficiency

Power is the rate at which energy is used or transferred and is measured in watts (W). The power of the immersion heater in this exercise is 400 W, indicating it transfers 400 joules of energy per second to the water. From an efficiency standpoint, not all the energy output from the heater is used effectively because only 80% of this energy is actually absorbed by the water.

Understanding energy efficiency is crucial because it measures how much input energy is effectively turned into useful output energy. It helps us decide how effectively a process uses energy, which was a critical part of the exercise in determining the real time needed for both heating and evaporation.

• Efficiency is calculated by dividing the useful energy output by the total energy input and multiplying by 100%.
• Improving energy efficiency often means less energy waste, leading to cost savings and reduced environmental impact.