Problem 114 - University Physics with Modern Physics

Question

Arod is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) throughout. One end is kept at \(0^{\circ} \mathrm{C}\) , and the other is brought into contact with a steam bath at \(100^{\circ} \mathrm{C}\) . The surface of the rod is insulated so that heat can flow only lengthwise along the rod. The cross-sectional area of the rod is \(2.50 \mathrm{cm}^{2},\) its length is 120 \(\mathrm{cm}\) , its thermal conductivity is \(380 \mathrm{W} / \mathrm{m} \cdot \mathrm{K},\) its density is \(1.00 \times 10^{4} \mathrm{kg} / \mathrm{m}^{3},\) and its specific heat is 520 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . Consider a short cylindrical element of the rod 1.00 \(\mathrm{cm}\) in length. (a) If the temperature gradient at the cooler end of this element is 140 \(\mathrm{C}^{\circ} / \mathrm{m}\) , how many joules of heat energy flow across this end per second? (b) If the average temperature of the element is increasing at the rate of 0.250 \(\mathrm{C} \%\) /s, what is the temperature gradient at the other end of the element?

Step-by-Step Solution

Answer

(a) 13.3 J/s (b) 105.9 C°/m

Step 1: Calculate Heat Flow

The heat flow across the cooler end due to the temperature gradient can be calculated using Fourier's law of heat conduction: \[ q = -kA \frac{dT}{dx} \] where \( q \) is the heat flow rate in watts, \( k = 380 \, \mathrm{W/m \cdot K} \) is the thermal conductivity, \( A = 2.50 \, \mathrm{cm^2} = 2.50 \times 10^{-4} \, \mathrm{m^2} \) is the cross-sectional area, and \( \frac{dT}{dx} = 140 \, \mathrm{C^{\circ}/m} \) is the temperature gradient. Substituting these values, we get: \[ q = -(380)(2.50 \times 10^{-4})(140) \] \[ q = -13.3 \, \mathrm{W} \] The negative sign denotes that heat flows in the direction of decreasing temperature.

Step 2: Determine Heat Energy Storage Change

The rate of change of internal energy in the rod element due to increasing temperature can be determined by: \[ \dot{Q}_{\text{stored}} = mc \frac{dT}{dt} \] where \( m = \rho V = \rho A \Delta x = (1.00 \times 10^4)(2.50 \times 10^{-4})(0.01) \) is the mass of the element, \( c = 520 \, \mathrm{J/kg \cdot K} \) is the specific heat, and \( \frac{dT}{dt} = 0.250 \, \mathrm{C^{\circ}/s} \) is the rate of temperature change. This gives: \[ \dot{Q}_{\text{stored}} = (1.00 \times 10^4)(2.50 \times 10^{-4})(0.01)(520)(0.250) \] \[ \dot{Q}_{\text{stored}} = 3.25 \, \mathrm{W} \]

Step 3: Calculate Heat Flow at Other End

The net heat flowing into the element must equal the rate of internal energy change plus the heat flowing out of the other end. Let the heat flow out of the hotter end be \( q_h \). Thus, \[ q_h = q + \dot{Q}_{\text{stored}} = -13.3 + 3.25 = -10.05 \, \mathrm{W} \]

Step 4: Find Temperature Gradient at Other End

Using Fourier's law for the hotter end cross-section, \( q_h = -kA \frac{dT_h}{dx} \), solve for \( \frac{dT_h}{dx} \): \[ \frac{dT_h}{dx} = \frac{-q_h}{kA} = \frac{10.05}{380 \times 2.50 \times 10^{-4}} = 105.9 \, \mathrm{C^{\circ}/m} \]

Key Concepts

Fourier's Law of Heat Conduction

Fourier's law of heat conduction is a fundamental principle in thermodynamics that describes the flow of heat in materials. It establishes a relationship between the heat transfer rate, the material's thermal conductivity, and the temperature gradient across the material. According to Fourier's law, the heat flow rate (\( q \)) is proportional to the negative gradient of temperature and the cross-sectional area through which heat flows. Mathematically, it can be expressed as \( q = -kA \frac{dT}{dx} \).
- \( k \) symbolizes the thermal conductivity of the material.
- \( A \) is the cross-sectional area.
- \( \frac{dT}{dx} \) represents the temperature gradient.
The negative sign indicates heat flows from high to low temperature.

Fourier's law is instrumental when calculating how much heat energy traverses a particular area of a substance over a specific period. Students often use this law to determine the heat flow in experiments involving rods or plates, especially when studying thermal insulation and heating mechanisms in various materials.

Thermal Conductivity

Thermal conductivity is a property of materials that measures their ability to conduct heat. It is denoted by the symbol \( k \) and is typically measured in watts per meter-kelvin (\( \mathrm{W/m \cdot K} \)). A high thermal conductivity indicates that the material can efficiently transfer heat, while a low thermal conductivity signifies that the material is a good insulator.
Materials like metals often have high thermal conductivities, making them ideal for applications requiring quick heat dissipation, such as in heat sinks or cooking ware. On the other hand, materials like wood or styrofoam have low thermal conductivity and are used where insulation is required.

Understanding thermal conductivity is crucial for designing systems where heat management is critical, such as in cooling systems for electronics, constructing energy-efficient buildings, and manufacturing thermal wear. In laboratory exercises, measuring thermal conductivity helps in identifying material properties and predicting how they'll perform under thermal stress. By knowing a rod's thermal conductivity, we can apply Fourier's law to calculate heat transfer through the rod, as shown in the example above.

Temperature Gradient

The temperature gradient is the rate of temperature change over a specific distance in a given direction and is denoted as \( \frac{dT}{dx} \). It is expressed in degrees per meter (\( \mathrm{C^{\circ}/m} \)) and is a pivotal factor in heat conduction calculations according to Fourier's law.

The temperature gradient influences how heat flows within materials. A steeper gradient means a larger change in temperature over distance, resulting in a greater heat flow rate across the material. In practical applications, this concept helps engineers design systems where specific temperature distributions need to be achieved, such as in temperature-controlled environments or industrial processes.

• High temperature gradients indicate quick heat transfers.
• Low temperature gradients suggest slower heat movement.

The exercise demonstrates how varying temperature gradients at different rod ends impact thermal flow rates. By calculating these gradients and corresponding heat transfers, students learn to analyze and predict how heat will disperse in real-world scenarios.

Specific Heat Capacity

Specific heat capacity (\( c \)) is a critical concept in thermodynamics, referring to the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or Kelvin). It is usually measured in joules per kilogram-kelvin (\( \mathrm{J/kg \cdot K} \)).

Materials with a high specific heat capacity can absorb more heat without experiencing a significant temperature increase, making them useful in maintaining stable temperatures. On the contrary, substances with low specific heat capacity heat up quickly but don't hold the temperature well. Water, for instance, has a high specific heat capacity, making it excellent for temperature regulation in natural and engineered systems.

In our exercise, the specific heat capacity of the rod informs us how much energy is absorbed as its temperature changes over time. The calculation of energy storage in the rod segment due to temperature rise is an application of specific heat capacity in real-world problems. It shows how different materials respond over time when subjected to thermal changes and helps predict their behavior in heating or cooling applications.