Problem 67 - College Physics

Question

Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawback is that energy is wasted because the tank loses heat when it is not in use, and you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is \(2.5 \mathrm{gal} / \mathrm{min}(9.46 \mathrm{~L} / \mathrm{min})\) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

Step-by-Step Solution

Answer

The rate of heat input required is approximately 25.78 kW.

Step 1: Understand the Problem

We need to find the rate of heat input required to heat water from 50°F (10°C) to 120°F (49°C) at a flow rate of 9.46 L/min using an on-demand heater. The heat capacity of water and the temperature change will be important factors.

Step 2: Determine Relevant Equations

The formula to calculate the heat energy required is \( Q = mc\Delta T \), where \( Q \) is heat energy, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change. The specific heat capacity of water is 4.186 J/g°C.

Step 3: Convert Flow Rate to Mass Flow Rate

First, convert the flow rate from liters per minute to grams per minute. Since 1 L of water is approximately 1000 g, the mass flow rate \( \dot{m} \) is \( 9.46 \, \text{L/min} \times 1000 \, \text{g/L} = 9460 \, \text{g/min} \).

Step 4: Calculate Temperature Change

Convert temperature from Celsius: initial temperature \( T_i = 10^{\circ} \text{C} \) and final temperature \( T_f = 49^{\circ} \text{C} \). The temperature change \( \Delta T \) is \( T_f - T_i = 49 - 10 = 39^{\circ} \text{C} \).

Step 5: Calculate Heat Required per Minute

Using the equation \( Q = mc\Delta T \), calculate the rate of heat needed per minute. \( Q = 9460 \, \text{g} \times 4.186 \, \text{J/g°C} \times 39^{\circ} \text{C} \).

Step 6: Compute Total Heat Input

Calculate \( Q \). The computation is \( Q = 9460 \times 4.186 \times 39 = 1,546,607.64 \, \text{J/min} \).

Step 7: Convert to Standard Power Units

Convert joules per minute to watts (1 watt = 1 joule/second). Since there are 60 seconds in a minute, \( 1,546,607.64 \, \text{J/min} \) is equal to \( \frac{1,546,607.64}{60} \approx 25,776.79 \, \text{W} \) (or approximately 25.78 kW).

Key Concepts

Heat Capacity of Water

Water's heat capacity is an essential factor in understanding how much heat energy is needed to alter its temperature. This property, known as specific heat capacity, is quite high for water, which means water requires a significant amount of energy to change its temperature.
In simple terms, heat capacity can be described as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is known to be 4.186 Joules per gram per degree Celsius (J/g°C).
• High heat capacity means that water can absorb a lot of heat before its temperature changes significantly.
• This property is why water is effective in regulating temperature in climate and in everyday appliances like heaters.

Understanding this concept is crucial when analyzing systems such as on-demand water heaters, where precise temperature control is necessary.

Temperature Change

Temperature change refers to the difference between the initial and final temperature of a substance when it is heated or cooled. In the context of our exercise, this plays a crucial role because it affects the amount of energy needed for heating.
The calculation is straightforward: \[\Delta T = T_f - T_i\]

Where:
• \( T_i \) is the initial temperature of the water (e.g., 10°C).
• \( T_f \) is the final temperature (e.g., 49°C).
• \( \Delta T \) is the temperature change, which in this case is \( 49°C - 10°C = 39°C \).
The more the temperature change, the more energy is needed, which is why precise calculation is essential for designing efficient heating systems.

Specific Heat Capacity

Specific heat capacity is the amount of heat per unit mass required to raise the temperature by one degree Celsius. This property varies among different substances and is a key parameter for designing heating and cooling systems.
Water, having a specific heat capacity of 4.186 J/g°C, implies it requires 4.186 Joules of energy to raise 1 gram of water by 1 degree Celsius.

Here's why understanding this is important:

• Water's high specific heat capacity helps in efficiently managing temperature changes.
• It is a major factor in the energy calculations for systems like on-demand water heaters.
• Knowing this can help optimize the design of such systems for energy conservation and effective performance.
  

In practical applications, specific heat capacity helps engineers determine how much energy will be needed for heating purposes.

Flow Rate Conversion

Flow rate conversion is important for calculating how much water is moving through a system at any given time, and hence, how much energy is required to heat it.
To effectively use metrics in your calculations, converting units is essential. For example, converting the flow rate from liters per minute to grams per minute involves knowing that 1 liter of water is approximately equivalent to 1000 grams.

For instance, if you have a flow rate of 9.46 L/min, convert it to grams like this:

• Multiply by 1000 to get grams: \( 9.46 \, \text{L/min} \times 1000 \, \text{g/L} = 9460 \, \text{g/min} \).

This conversion allows you to use the water mass flow rate in energy calculations, which is essential when assessing the heating requirements with an on-demand system.