Problem 20 - College 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 \(200.0 \mathrm{~W}\) electric immersion heater in \(0.320 \mathrm{~kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required if all of the heater's power goes into heating the water?

Step-by-Step Solution

Answer

(a) 80275.2 J; (b) 401.376 seconds.

Step 1: Determine the heat needed

To find the amount of heat required, we use the formula for heat change: \[ Q = mc\Delta T \]where \( Q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.Given:- Mass of water \( m = 0.320 \text{ kg} \)- Specific heat capacity of water \( c = 4.186 \text{ J/g}^\circ \text{C} = 4186 \text{ J/kg}^\circ \text{C} \)- Temperature change \( \Delta T = 80.0 - 20.0 = 60.0^\circ \text{C} \)Substitute these values into the formula:\[ Q = 0.320 \times 4186 \times 60.0 = 80275.2 \text{ J} \]

Step 2: Calculate the time required using power

To find the time required, we use the relationship between power, energy, and time:\[ P = \frac{Q}{t} \]Where \( P \) is power, \( Q \) is energy, and \( t \) is time.Rearrange for \( t \):\[ t = \frac{Q}{P} \]Given:- Power \( P = 200.0 \text{ W} \)- Energy \( Q = 80275.2 \text{ J} \)Substitute these into the formula:\[ t = \frac{80275.2}{200.0} = 401.376 \text{ seconds} \]

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy required to change the temperature of a substance. It is a property that varies between different materials. For water, the specific heat capacity is high, meaning it requires more energy to raise its temperature compared to many other substances. In the exercise, the specific heat capacity of water is given as 4186 J/kg°C. This means that for every kilogram of water, 4186 joules are required to increase its temperature by 1 °C.
Some key points about specific heat capacity:

• It is typically measured in joules per kilogram per degree Celsius (J/kg°C).
• It explains why materials like water are good at storing and transferring heat.
• This concept is crucial in thermodynamics and is applied in various fields like cooking, heating systems, and climate studies.

Knowing the specific heat capacity of a material helps in calculating how much energy is needed for heating or cooling processes.

Energy Transfer

Energy transfer in thermodynamics usually involves the movement of energy from one system to another. In the case of the coffee-making process described, energy is transferred from the electrical heater to the water. This energy transfer causes the water's temperature to increase.
Here are some important aspects of energy transfer:

• Energy can be transferred through various methods, including conduction, convection, and radiation.
• In this exercise, conduction is the primary method of energy transfer, where heat flows directly from the heater into the water.
• Consistent application of energy results in a rise in water temperature, until the desired level is achieved.

Understanding energy transfer is vital for the efficient design of heating systems, thermal engines, and even simple everyday tasks like boiling water.

Power and Work

Power and work are closely linked in the field of physics. Power refers to the rate of doing work or transferring energy over time. In simple terms, it's a measure of how quickly energy is being used or transferred.
In our exercise, a 200 W heater is used, where watts (W) is a unit of power. This means the heater transfers 200 joules of energy per second into the water.

• Power is calculated as energy (work done) divided by time, represented by the formula: \( P = \frac{Q}{t} \).
• The effectiveness of heating appliances directly relates to their power rating - higher power ratings equate to faster energy transfer.
• Understanding power is useful in comparing the efficiency of different appliances and ensuring energy and cost efficiency.

This understanding helps you choose the right appliances for efficient energy use and optimal performance in everyday activities.

Heat Transfer Calculations

Heat transfer calculations are essential for determining how much energy is needed to change a substance's temperature. The base formula used for calculating heat energy is: \( Q = mc\Delta T \), where:

• \( Q \) is the heat energy transferred.
• \( m \) is the mass of the substance.
• \( c \) is the specific heat capacity of the substance.
• \( \Delta T \) is the change in temperature.

In the exercise, by applying the formula, we found that 80275.2 J of heat energy is needed to heat 0.320 kg of water from 20 °C to 80 °C.
Heat transfer calculations not only inform us how much energy we need but also help us understand the efficiency of heating systems, the choice of materials for thermal storage, and the environmental impacts.
Accurate calculations ensure optimal energy use, reduce costs, and contribute to environmental conservation efforts.