Problem 28 - An Introduction to Thermal Physics

Question

Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600 -watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.

Step-by-Step Solution

Answer

It takes approximately 2.18 minutes to boil the water. All energy is assumed to be used in heating.

Step 1: Understanding the Problem

We need to estimate the time required to bring a cup of water to boiling using a 600-watt microwave. Assume the microwave's energy efficiency is 100%, meaning all energy heats the water.

Step 2: Identify Known Values

Given: Microwave power = 600 watts, specific heat of water = 4.18 J/g°C. Assume 1 cup of water = 250 grams (as 1 mL of water weighs about 1 gram), and the initial temperature is room temperature, say 25°C. Water boils at 100°C.

Step 3: Calculate Energy Needed

The energy needed (in joules) is calculated with the formula: \[ E = m imes c imes \Delta T \] where \( m = 250 \) grams, \( c = 4.18 \) J/g°C, and \( \Delta T = 100°C - 25°C = 75°C \).Plug values into equation:\[ E = 250 imes 4.18 imes 75 \approx 78,375 \text{ J} \].

Step 4: Relate Energy to Time

Microwave power tells us that 600 J of energy is transferred per second. To find the time \( t \) (in seconds), use the formula:\[ t = \frac{E}{P} \]where \( E = 78,375 \text{ J} \) and \( P = 600 \text{ W} \). Compute:\[ t = \frac{78,375}{600} \approx 130.625 \text{ seconds} \].

Step 5: Convert Time to Minutes

Convert the time from seconds to minutes:\[ \text{Minutes} = \frac{130.625}{60} \approx 2.18 \text{ minutes} \].

Step 6: Explain Why No Heat is Involved

The question might refer to the fact that heat loss isn't needed in the analysis due to the assumption that all microwave energy directly contributes to heating the water. This is why there's no need to account for lost heat in our calculation.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding how substances heat up and cool down. It represents the amount of energy needed to raise the temperature of 1 gram of a substance by 1°C. In our exercise, we used the specific heat capacity of water, which is 4.18 J/g°C. This means that it takes 4.18 joules of energy to increase the temperature of one gram of water by one degree Celsius.

Specific heat capacity is a property specific to each material and helps explain why different substances heat up at different rates. For example:

• Water requires more energy to change its temperature compared to metals, which have lower specific heat capacities.
• This is why water heats up more slowly than frying oil or metals.
• Understanding specific heat capacity helps in calculating the energy needed for heating processes in various applications.

In microwave heating, knowing the specific heat capacity of the substance being heated, like water, enables accurate calculations of energy requirements.

Energy Transfer

Energy transfer in the context of microwave heating involves the conversion of microwave radiation into thermal energy, which raises the temperature of the food or liquid. In this exercise, the microwave oven emits electromagnetic waves, which are absorbed by water molecules, causing them to vibrate and produce heat.

The energy transfer process in microwaves happens very quickly because microwaves penetrate the substance and directly transfer energy to the molecules. Here are some important points about energy transfer in microwaves:

• Microwave energy is very efficient as it directly targets water molecules within the food or drink.
• Unlike traditional cooking that heats food from the outside, microwaves heat from the inside out.
• The efficiency of energy transfer in a microwave depends on its wattage; in our example, the microwave is 600 watts, meaning it transfers 600 joules of energy per second.

By understanding energy transfer, we can estimate how long it will take to heat food or liquid to a desired temperature.

Heat Calculations

Heat calculations are essential when determining how much energy is needed to raise the temperature of an object or substance. In our problem, we needed to calculate the total energy required to bring water from room temperature (25°C) to its boiling point (100°C).

We used the heat energy formula: \[ E = m \times c \times \Delta T \]where:

• \( E \) is the energy in joules required for heating,
• \( m \) is the mass of the water (250 grams for 1 cup),
• \( c \) is the specific heat capacity of water (4.18 J/g°C), and
• \( \Delta T \) is the change in temperature (75°C).

In this calculation, we found that 78,375 joules of energy were needed. By knowing the power of the microwave (600 watts), we could calculate the time it takes to provide this energy through:\[ t = \frac{E}{P} \]This helped us estimate that it takes about 2.18 minutes to boil the water in the microwave, showing the practical application of heat calculations in everyday tasks.