Problem 24 - University Physics with Modern Physics

Question

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200-W electric immersion heater in 0.320 kg of water. (a) How much heat must be added to the water to raise its temperature from 20.0\(^\circ\)C to 80.0\(^\circ\)C? (b) How much time is required? Assume that all of the heater’s power goes into heating the water.

Step-by-Step Solution

Answer

(a) 80,179.2 J; (b) 400.896 seconds.

Step 1: Identify What is Known

We know the initial temperature \( T_i = 20.0^\circ C \), the final temperature \( T_f = 80.0^\ C \), and the mass of the water \( m = 0.320 \, \text{kg} \). The specific heat capacity of water \( c \) is approximately \( 4,186 \, \text{J/(kg} \, \text{°C)} \).

Step 2: Calculate the Heat Required (Q)

Use the formula for heat transfer: \( Q = mc\Delta T \). The temperature change \( \Delta T \) is \( T_f - T_i = 80 - 20 = 60 \, \text{°C} \). Therefore, \[ Q = 0.320 \times 4,186 \times 60 = 80,179.2 \, \text{J}. \] This is the heat required to raise the water's temperature.

Step 3: Calculate the Time Required

Given the heater power \( P = 200 \, \text{W} \), and knowing that power is energy per unit time (\( P = \frac{Q}{t} \)), rearrange to find time \( t \). Thus, \( t = \frac{Q}{P} = \frac{80,179.2}{200} \). Calculating gives \[ t = 400.896 \, \text{seconds}. \]

Step 4: Review and Confirm the Solution

Ensure calculations for both parts (a) and (b) align with the expected physics principles. The result indicates both the heat required and the time needed when using the given heater.

Key Concepts

Specific Heat Capacity

Understanding specific heat capacity is essential when dealing with thermal processes, like heating water. The specific heat capacity is a measure of how much heat energy a substance can absorb before its temperature changes. In mathematical terms, it's denoted as \( c \) and is defined as the amount of heat (Q) required to raise the temperature of 1 kg of a substance by 1°C.
Water has a relatively high specific heat capacity, approximately 4,186 J/(kg°C). This means water can absorb a lot of heat without its temperature rising too quickly.
For students working with heat calculations, remember that higher specific heat capacities mean a substance can store more energy per kilogram, enhancing its ability to resist temperature change. This is why water, with its high specific heat, is often used in heating and cooling applications.

Heat Transfer

Heat transfer is the movement of thermal energy from one place to another, and it's a fundamental principle in physics. When heating water with an electric heater, heat transfer occurs from the heater to the water, raising its temperature.
Three main methods facilitate heat transfer:

• Conduction: Heat travels through a substance without any movement of the substance itself. In our example, heat transfers from the electric heater to the water through direct contact.
• Convection: This involves the movement of fluid (gas or liquid) carrying heat with it. In a pot of water, convection currents distribute heat evenly throughout the liquid.
• Radiation: Heat transfer through electromagnetic waves, like sunlight warming your skin.

In heating the water from 20°C to 80°C as described in the exercise, the formula \( Q = mc\Delta T \) calculates the energy required, with \( \Delta T \) representing the temperature change. This calculation accurately predicts the energy needed to achieve the desired temperature increase.

Electric Heaters

Electric heaters are modern solutions for converting electrical energy into heat energy efficiently. In the exercise, a 200-Watt immersion heater is used to warm water. Such devices are simple yet effective because they convert almost all their electrical input into heat, making them very practical for specific tasks like heating water.
The power of an electric heater, stated in watts, indicates how much energy it uses per second. In the problem, the 200-Watt heater applies this energy directly to the water, with all the energy from the heater transferred to the water.
To calculate how long it takes the heater to warm up the water to a specific temperature, the equation \( t = \frac{Q}{P} \) is used, where \( Q \) is the energy required and \( P \) is the power of the heater. This allows you to determine precisely how efficient the heater is for your needs, making electric heaters a reliable choice for quick and effective heat transfer.