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The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 52-gal \(\left(1.00 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\right)\) unit starts with cold water at \(11^{\circ} \mathrm{C}\) and delivers hot water at \(53^{\circ} \mathrm{C}\). The unit is electric and utilizes a resistance heater \((120 \mathrm{~V} \mathrm{ac}, 3.0 \Omega)\) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.

Short Answer

Expert verified
The recovery time is approximately 2 hours.

Step by step solution

01

Convert gallons to cubic meters

Convert the volume of water from gallons to cubic meters using the conversion factor. We are given a 52-gal unit, and 1 gallon is equivalent to \(3.79 \times 10^{-3} \mathrm{~m}^{3}\). Thus, the volume \( V \) in cubic meters is calculated as: \[ V = 52 \times 3.79 \times 10^{-3} \approx 0.19708 \; \mathrm{m}^{3} \]
02

Calculate the mass of water

Calculate the mass of water using the density of water (\(1000 \mathrm{~kg/m^3}\)). The formula for mass \( m \) is: \[ m = \text{density} \times \text{volume} = 1000 \times 0.19708 \approx 197.08 \; \mathrm{kg} \]
03

Determine energy required to heat the water

The energy required \(Q\) to heat the water is given by the formula: \[ Q = mc\Delta T \]where \(c = 4186 \; \mathrm{J/kg^{ullet}C}\) is the specific heat capacity of water, and \(\Delta T\) is the temperature change from \(11^{\circ}\mathrm{C}\) to \(53^{\circ}\mathrm{C}\), which is \(42^{\circ}\mathrm{C}\). Substituting the values, \[ Q = 197.08 \times 4186 \times 42 \approx 34572564.64 \; \mathrm{J} \]
04

Calculate power supplied by the heater

Using the resistance heater's specifications (\(120 \mathrm{~V}, 3.0 \Omega\)), calculate the power \(P\) using the formula: \[ P = \frac{V^2}{R} \]where \(V = 120 \mathrm{~V}\) is the voltage and \(R = 3.0 \Omega\) is the resistance. Therefore, \[ P = \frac{120^2}{3} = 4800 \; \mathrm{W} \]
05

Calculate the recovery time

The recovery time \(t\) is given by \(t = \frac{Q}{P}\), which is the energy required divided by the power output. Substitute the values obtained: \[ t = \frac{34572564.64}{4800} \approx 7202.6 \; \mathrm{seconds} \] Convert seconds to hours: \[ t = \frac{7202.6}{3600} \approx 2.00 \; \mathrm{hours} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Energy Calculations
To determine the recovery time of a hot water heater, we first need to calculate the thermal energy required to heat the water. Thermal energy refers to the energy that is transferred to raise the water's temperature. In this scenario, we use the formula:\[ Q = mc\Delta T \]where
  • \( Q \) is the thermal energy in joules (J),
  • \( m \) is the mass of the water in kilograms (kg),
  • \( c \) is the specific heat capacity of the water (more on this in the next section), and
  • \( \Delta T \) is the change in temperature in degrees Celsius (°C).
Calculate the change in temperature by subtracting the initial temperature (\( 11^{\circ} \mathrm{C} \)) from the desired temperature (\( 53^{\circ} \mathrm{C} \)), resulting in a \( \Delta T \) of \( 42^{\circ} \mathrm{C} \). We already know the mass and specific heat capacity. Given this information, you can find the thermal energy required.
Specific Heat Capacity
Specific heat capacity is a crucial concept when it comes to heating substances, especially water. Specific heat capacity is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. For water, this value is relatively high at \( 4186 \, \mathrm{J/kg^{\bullet}C} \), meaning water requires a significant amount of energy to change its temperature.
This property makes water an excellent substance for maintaining temperatures in heating systems. Since the specific heat capacity of water is consistent, you can reliably calculate the energy needed to achieve a desired temperature change.In thermal energy calculations, always ensure to measure or know the substance's specific heat capacity you are working with to make accurate predictions or solutions.
Electric Resistance Heating
Electric resistance heating involves an electric current passing through a resistance, converting electrical energy into heat. This principle is used in electric water heaters to elevate water to a desired temperature. The efficiency and power of this heating system depend upon the voltage supplied and the resistance of the heating element.To assess the power supplied by the heater, you can apply the formula:\[ P = \frac{V^2}{R} \]where
  • \( P \) is the power in watts (W),
  • \( V \) is the voltage in volts (V),
  • \( R \) is the resistance in ohms (\( \Omega \)).
In the given exercise, with a voltage of \( 120 \, \mathrm{V} \) and a resistance of \( 3.0 \, \Omega \), the power is determined to be \( 4800 \, \mathrm{W} \). Larger power output implies quicker heating; thus, a factor that can significantly influence recovery time. Always consider these parameters early to ensure efficient system design and operation.

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Most popular questions from this chapter

A wire has a resistance of \(21.0 \Omega\). It is melted down, and from the same volume of metal a new wire is made that is three times longer than the original wire. What is the resistance of the new wire?

A portable CD player operates with a voltage of \(4.5 \mathrm{~V},\) and its power usage is \(0.11 \mathrm{~W}\). Find the current in the player.

A \(75.0-\Omega\) and a \(45.0-\Omega\) resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is \(0.294 \mathrm{~A}\). When the \(45.0-\Omega\) resistor is disconnected, the current from the battery drops to \(0.116 \mathrm{~A}\). Determine (a) the emf and (b) the internal resistance of the battery.

A galvanometer has a full-scale current of \(0.100 \mathrm{~mA}\) and a coil resistance of \(50.0 \Omega\). This instrument is used with a shunt resistor to form an ammeter that will register full scale for a current of \(60.0 \mathrm{~mA}\). Determine the resistance of the shunt resistor.

In Section \(12.3\) it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is \(125 \Omega\) when its temperature is \(20.0^{\circ} \mathrm{C}\). The wire is then immersed in boiling chlorine, and the resistance drops to \(99.6 \Omega\). The temperature coefficient of resistivity of platinum is \(\alpha=3.72 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\). What is the temnerature of the boiling chlorine?

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