Problem 83 - Fundamentals of Physics

Question

An electric immersion heater normally takes 100 min to bring cold water in a well-insulated container to a certain temperature, after which a thermostat switches the heater off. One day the line voltage is reduced by \(6.00 \%\) because of a laboratory overload. How long does heating the water now take? Assume that the resistance of the heating element does not change.

Step-by-Step Solution

Answer

The heating now takes approximately 113.17 minutes.

Step 1: Understanding the Problem

We are given that an immersion heater normally takes 100 minutes to heat water. One day, the voltage is reduced by \(6.00\%\). We need to calculate the new time it will take to heat the water given that the resistance of the heater remains constant.

Step 2: Identify Relationship Between Power, Voltage, and Time

Power \(P\) is related to energy \(E\) and time \(t\) by the formula \(P = \frac{E}{t}\). The power consumed by a circuit is also determined by the voltage \(V\) and resistance \(R\) as \(P = \frac{V^2}{R}\). Since \(R\) is constant, changes in \(V\) will affect \(P\) and, consequently, \(t\).

Step 3: Calculate Original Power

The original power is expressed as \( P_{ ext{original}} = \frac{V^2}{R} \). The time taken for heating is \( t_{ ext{original}} = 100 \) minutes, so \( E = P_{ ext{original}} \times t_{ ext{original}} \).

Step 4: Calculate New Voltage and Power

If voltage is reduced by \(6.00\%\), the new voltage \( V_{ ext{new}} = 0.94V \). The new power \( P_{ ext{new}} = \frac{(0.94V)^2}{R} = 0.8836 \times \frac{V^2}{R} = 0.8836P_{ ext{original}} \).

Step 5: Relate Power and Time After Voltage Change

The energy required to heat the water remains the same as before, so we can set \( E = P_{ ext{new}} \times t_{ ext{new}} \). Therefore, \( t_{ ext{new}} = \frac{E}{P_{ ext{new}}} = \frac{E}{0.8836P_{ ext{original}}} \). Simplifying gives \( t_{ ext{new}} = \frac{t_{ ext{original}}}{0.8836} \).

Step 6: Compute the New Heating Time

Substitute \( t_{ ext{original}} = 100 \) minutes: \( t_{ ext{new}} = \frac{100}{0.8836} \approx 113.17 \) minutes.

Key Concepts

Power Calculations

Power calculations are an essential part of understanding how electric circuits operate, especially when working with devices like heating elements. The power in a circuit is the rate at which energy is used. It's calculated using the formula \( P = \frac{E}{t} \), where \( P \) represents power, \( E \) is energy, and \( t \) is time. For electric devices, power can also be expressed as \( P = \frac{V^2}{R} \). Here, \( V \) stands for voltage, \( R \) is resistance.

• Power allows us to understand how much energy is consumed over time. High power means more energy usage and faster operation, whereas low power means slower operation.
• In our exercise, the heater requires a certain power to bring water to the desired temperature in a specific amount of time.
• When voltage changes, it directly impacts power, altering how quickly an electric device can perform its function.

Understanding these calculations is crucial for determining how changes in a circuit, like voltage drop, will affect overall performance.

Voltage Reduction

Voltage reduction in an electric circuit can significantly affect the performance of devices such as heating elements. Voltage is the potential difference that drives current through a circuit. In our scenario, a \(6\%\) reduction in voltage affects how efficiently the heater operates.

• With lower voltage, the power, calculated as \( P = \frac{V^2}{R} \), decreases. Since power is proportional to the square of the voltage, even a small voltage reduction can lead to a noticeable drop in power.
• This reduced power output means that the heater takes longer to reach the same level of performance, which in our case, is bringing water to a certain temperature.
• Voltage reduction can be caused by various factors like lab overloads and can impact not just heating times but also the functionality of other sensitive equipment.

Being aware of how voltage drops affect performance is important for maintaining efficiency and effectiveness in electric circuits.

Heating Elements

Heating elements are devices used to convert electrical energy into heat. They are made from materials with high electrical resistance, allowing them to generate heat efficiently when current passes through.

• They are used in a wide range of appliances, from water heaters to ovens and even irons.
• The efficiency of a heating element is affected by the power available, which is dictated by the voltage and the intrinsic resistance of the element itself.
• In the given exercise, since the resistance remains constant, the heating element's power adapts solely based on changes in the voltage.

When the voltage decreases, as shown in our problem, the time taken to achieve the desired heating increases because the power output of the heating element is reduced. Understanding the interplay between voltage, resistance, and power in heating elements helps in optimizing their use across various applications.