91Ó°ÊÓ

45\. Building a Water Heater The nickel-chromium alloy Nichrome has a resistivity of about \(10^{-6} \Omega-\mathrm{m} .\) Suppose you want to build a small heater out of a coil of Nichrome wire and a \(6 \mathrm{~V}\) battery in order to heat \(30 \mathrm{ml}\) of water from a temperature of \(20 \mathrm{C}\) to \(40 \mathrm{C}\) in \(1 \mathrm{~min}\). Assume the battery has negligible internal resistance. (a) How much heat energy (in joules) do you need to do this? (b) How much power (in watts) do you need to do it in the time indicated? (c) What resistance should your Nichrome coil have in order to produce this much power in heat? (d) Can you create a coil having these properties? (Hint: Can you find a plausible length and cross-sectional area for your wire that will give you the resistance you need?) (e) If the internal resistance of the battery were \(1 / 3 \Omega\), how would it affect your calculation? (Only explain what you would have to do; don't recalculate the size of your coil.)

Step by step solution

01

Calculate Required Heat Energy

First, determine the amount of heat energy needed to heat the water. Use the formula for heat energy: \[ Q = mc\Delta T \]where: \( Q \) = heat energy (in joules)\( m \) = mass of water (in kg)\( c \) = specific heat capacity of water (\( 4186 \mathrm{~J/kg-K} \))\( \Delta T \) = change in temperature (in K)Convert the volume of water to mass (since 1 ml of water has a mass of about 1 g): \( m = 30 \mathrm{ml} = 30 \mathrm{g} = 0.03 \mathrm{kg} \)The change in temperature is: \( \Delta T = 40^\text{C} - 20^\text{C} = 20^\text{C} = 20 \text{K} \)Calculate the heat energy: \[ Q = 0.03 \times 4186 \times 20 = 2511.6 \text{~J} \]

02

Calculate Required Power

Determine the power required to produce this heat energy in the given time. Use the formula: \[ P = \frac{Q}{t} \]where: \( P \) = power (in watts)\( Q \) = heat energy (in joules)\( t \) = time (in seconds)Given time is 1 minute: \( t = 60 \text{~s} \)Calculate the power: \[ P = \frac{2511.6}{60} = 41.86 \text{~W} \]

03

Calculate Required Resistance

Use Ohm's law and the power formula to find the resistance. The formulas are: \[ V = IR \quad \text{and} \quad P = \frac{V^2}{R} \]Rearrange to solve for the resistance \( R \): \[ R = \frac{V^2}{P} \]Given voltage \( V = 6 \text{~V} \) and power \( P = 41.86 \text{~W} \): \[ R = \frac{6^2}{41.86} = \frac{36}{41.86} \approx 0.86 \Omega \]

04

Determine Feasibility of Coil

Check if it is possible to create a Nichrome coil with the required resistance. Use the resistivity formula: \[ R = \rho \frac{L}{A} \]where: \( \rho \) = resistivity of Nichrome (\( 10^{-6} \Omega\text{~m} \))\( L \) = length of the wire (in m)\( A \) = cross-sectional area (in \( \mathrm{m}^2 \))Rework the formula to solve for \( L \) and \( A \). For a plausible coil, assume a wire diameter and calculate the plausible length.

05

Effect of Internal Resistance of Battery

If the battery has considerable internal resistance (e.g., \( 1/3 \Omega \)), it will affect the total resistance in the circuit. You need to include the internal resistance in the total resistance: \[ R_{total} = R_{coil} + R_{internal} \]Recalculate the resistance of the Nichrome coil with the internal resistance factored in to achieve the same power output.

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Key Concepts

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

Nichrome Wire Resistivity

Nichrome is a blend of nickel and chromium. It's popular in heating elements due to its high resistivity and ability to withstand high temperatures. The resistivity of Nichrome can be represented by the symbol \( \rho \), and for Nichrome, it's approximately \( 10^{-6} \, \Omega \text{m} \). Resistivity is the property that quantifies how strongly a material opposes the flow of electric current. The higher the resistivity, the harder it is for current to flow through the material. Understanding this is crucial when designing electric heaters, as it helps in calculating the necessary dimensions and lengths of Nichrome wire to achieve the desired resistance.

Heat Energy Calculation

To determine how much heat energy is required to raise the temperature of a substance, we use the formula: \( Q = mc\Delta T \). Here:

• \( Q \) is the heat energy in joules (J).
• \( m \) is the mass of the substance in kilograms (kg).
• \( c \) is the specific heat capacity of the substance in \( \text{J/kg} \cdot \text{K} \). For water, \( c \) is \( 4186 \text{~J/kg} \cdot \text{K} \).
• \( \Delta T \) is the change in temperature in Celsius (C) or Kelvin (K), since \( \Delta T \) is the same in both units.

For our water heater, we wanted to heat 30 ml of water from 20°C to 40°C within 1 minute. Thus:

• Volume of water: 30 ml, which is 30 grams or 0.03 kg.
• Temperature change: \( 40 \text{C} - 20 \text{C} = 20 \text{C} \).

Using these values:
\( Q = 0.03 \times 4186 \times 20 = 2511.6 \text{~J} \). This equation tells us that 2511.6 Joules of heat energy is needed.

Ohm's Law

Ohm's Law is fundamental in understanding electrical circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The law is usually written as:
\( V = IR \), where:

• \( V \) is the voltage in volts (V).
• \( I \) is the current in amperes (A).
• \( R \) is the resistance in ohms (\( \Omega \)).

In our water heater example, we determined we needed a resistance of about 0.86 ohms using the formula \( P = \frac{V^2}{R} \). Given that we wanted to use a 6V battery and needed about 41.86 watts of power:
\( R = \frac{6^2}{41.86} = \frac{36}{41.86} \approx 0.86 \Omega \). Knowing this resistance, we could design the coil to meet this requirement.

Specific Heat Capacity of Water

The specific heat capacity is a property that tells us how much heat energy is needed to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). For water, the specific heat capacity is quite high, at \( 4186 \text{~J/kg} \cdot \text{K} \). This means it takes 4186 joules to heat one kilogram of water by one degree. In our problem:

• the water's mass was 0.03 kg.
• We wanted to increase the temperature by 20°C.

Plugging these values into the heat energy formula (\( Q = mc\Delta T \)), we calculated:
\( Q = 0.03 \times 4186 \times 20 = 2511.6 \text{~J} \). This shows how knowing the specific heat capacity of water allows us to calculate the total heat energy required accurately.