Problem 52 - Physics

Question

A heat lamp emits infrared radiation whose rms electric field is \(E_{\mathrm{rms}}=2800 \mathrm{~N} / \mathrm{C} .\) (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius \(4.0 \mathrm{~cm}\). What is the average power delivered to the leg? (c) The portion of the leg being radiated has a mass of \(0.28 \mathrm{~kg}\) and a specific heat capacity of \(3500 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) How long does it take to raise its temperature by \(2.0 \mathrm{C}^{\circ}\) ? Assume that there is no other heat transfer into or out of the portion of the leg being heated.

Step-by-Step Solution

Answer

(a) 10400 W/m^2; (b) 52 W; (c) ≈ 38 seconds.

Step 1: Calculate Average Intensity - Formula

The average intensity of the radiation can be calculated using the formula:\[ I = \frac{1}{2} c \epsilon_0 E_{\text{rms}}^2 \]where:- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \) m/s).- \( \epsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \) F/m.- \( E_{\text{rms}} \) is the root mean square of the electric field, given as \( 2800 \text{ N/C} \).

Step 2: Calculate Average Intensity - Substitution & Calculation

Substitute the given values into the formula:\[I = \frac{1}{2} \cdot (3 \times 10^8) \cdot (8.85 \times 10^{-12}) \cdot (2800)^2\]Calculating this gives:\[ I = 1.04 \times 10^{4} \text{ W/m}^2 \]

Step 3: Calculate Area of the Circle

The circular area on the leg can be calculated using the formula for the area of a circle:\[ A = \pi r^2 \]where:- \( r = 4.0 \) cm = 0.04 m. Substituting the radius gives:\[ A = \pi \times (0.04)^2 = 5.0 \times 10^{-3} \text{ m}^2 \]

Step 4: Calculate Average Power Delivered - Formula & Calculation

The average power delivered to the leg can be calculated using:\[P = I \cdot A \]Substitute \( I = 1.04 \times 10^{4} \text{ W/m}^2 \) and \( A = 5.0 \times 10^{-3} \text{ m}^2 \):\[ P = 1.04 \times 10^4 \times 5.0 \times 10^{-3} = 52 \text{ W} \]

Step 5: Determine Heat Required

The heat required to raise the leg's temperature can be calculated using the formula:\[ Q = m c \Delta T \]where:- \( m = 0.28 \) kg.- \( c = 3500 \) J/kg°C.- \( \Delta T = 2.0 \)°C.Substituting these values gives:\[ Q = 0.28 \times 3500 \times 2 = 1960 \text{ J} \]

Step 6: Calculate Time Required - Formula & Calculation

The time required to deliver this energy with constant power is:\[ t = \frac{Q}{P} \]With \( Q = 1960 \text{ J} \) and \( P = 52 \text{ W} \), we find:\[ t = \frac{1960}{52} \approx 37.7 \text{ seconds} \]

Key Concepts

Electric Field

Understanding the electric field is crucial when discussing infrared radiation. An electric field is a field around charged particles that exerts force on other charges within the field. The strength of this field is measured in newtons per coulomb (N/C). In the context of a heat lamp emitting infrared radiation, the **root mean square** (rms) value of the electric field is given. This rms value represents the effective intensity of the fluctuating electric field from the radiation, providing a means to calculate further electric-based properties. The rms electric field is crucial because it allows us to determine the average intensity of radiation, which, as seen in the original exercise, is 2800 N/C. This measure helps ascertain how strongly the field can interact with particles, thus affecting energy transfer.

Intensity of Radiation

The intensity of radiation quantifies the power per unit area received from a radiation source. It's an important concept as it gives us insight into how powerful the radiation source is. Mathematically, intensity is denoted by the formula:

• \[ I = \frac{1}{2} c \epsilon_0 E_{ ext{rms}}^2 \]

where

• \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \text{ m/s} \),
• \( \epsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \text{ F/m} \),
• \( E_{\text{rms}} \) is the root mean square of the electric field.

This formula derives from electromagnetic theory, where the intensity relates the field properties to energy density. In the example given, using the known values, the average intensity is calculated to be about \( 1.04 \times 10^{4} \text{ W/m}^2 \). This intensity reflects how much power the heat lamp delivers over a specified area, which is essential for heating calculations.

Specific Heat Capacity

Specific heat capacity is a material-specific property that indicates how much heat energy is required to raise the temperature of a unit mass of the substance by one degree Celsius (or Kelvin). In the given exercise, the specific heat capacity of the person's leg is assumed at \( 3500 \, \text{J/kg°C} \). This value suggests that every kilogram of the leg material needs 3500 joules of energy to increase its temperature by 1°C.
Specific heat capacity is fundamental when considering how substances absorb or release energy when their temperature changes. If a material has a high specific heat capacity, it can absorb a lot of heat before its temperature changes significantly. Conversely, a low specific heat results in a noticeable temperature change with less applied heat.

Thermal Energy Transfer

Thermal energy transfer denotes the process of heat movement from one body or system to another due to a temperature difference. In the context of the lamp and the exercise, thermal energy moves from the heat lamp's radiation to the leg. The amount of thermal energy transferred increases the temperature of the leg, calculated using the formula:

• \[ Q = mc\Delta T \]

where

• \( Q \) is the heat energy (in joules),
• \( m \) is the mass (in kilograms),
• \( c \) is the specific heat capacity (in \( \text{J/kg°C} \)),
• \( \Delta T \) is the change in temperature (in °C).

From the exercise, calculating using this formula shows that raising the leg's temperature requires 1960 J of energy. The time taken for this energy transfer depends on the power delivered by the radiation. Using the derived power, it's shown that about 37.7 seconds are needed to reach the desired temperature increment, illustrating the dynamic relationship between power, specific heat, and time.