Question

Two kilograms of a gas with molecular weight 28 are contained in a closed, rigid tank fitted with an electric resistor. The resistor draws a constant current of \(10 \mathrm{amp}\) at a voltage of \(12 \mathrm{~V}\) for \(10 \mathrm{~min}\). Measurements indicate that when equilibrium is reached, the temperature of the gas has increased by \(40.3^{\circ} \mathrm{C}\). Heat transfer to the surroundings is estimated to occur at a constant rate of \(20 \mathrm{~W}\). Assuming ideal gas behavior, determine an average value of the specific heat \(c_{p}\), in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), of the gas in this temperature interval based on the measured data.

Short answer

The average value of the specific heat (\[ c_p \]) is approximately \[0.074 \text{kJ/kg·K}\].

Step-by-step solution

Identify the given values

List the given information:- Mass of gas, - Molecular weight of gas, - Current through resistor, - Voltage across resistor, - Duration of current, - Temperature increase, - Heat transfer rate to surroundings.

Calculate work done by the resistor

Calculate the electrical work done by the resistor using the formula \[ W = V \times I \times t \] where V is the voltage, I is the current, and t is the time in seconds.Given: - Voltage, - Current, - Time.Work done, \[ W = 12 \times 10 \times (10 \times 60) = 7200 \text{ Joules} \] or \[ W = 7.2 \text{ kJ} \]

Calculate heat transfer to surroundings

Calculate the heat transfer to surroundings using the formula \[ Q_{\text{loss}} = \text{power} \times t \]Given: - Power, - Time.Heat transfer to surroundings, \[ Q_{\text{loss}} = 20 \times (10 \times 60) = 12000 \text{ Joules} \] or \[ Q_{\text{loss}} = 1.2 \text{ kJ} \]

Calculate net heat added to the gas

Calculate the net heat added to the gas. Use the formula \[ Q_{\text{gas}} = W - Q_{\text{loss}} \] Substitute the values from the previous steps.Net heat added to gas, \[ Q_{\text{gas}} = 7.2 - 1.2 = 6.0 \text{ kJ} \]

Calculate temperature increase

Temperature increase is already given as \[ \triangle T = 40.3^{\text{C}} = 40.3 \text{K} \] (conversion from C to K is not needed here since we're working with temperature differences).

Determine the average specific heat

Use the formula \[ Q = mc_p \triangle T \] where m is the mass of gas. Solve for specific heat, given Q, m, and \[ \triangle T, \]Substitute the known values, \[ c_p = \frac{Q}{m \times \triangle T} = \frac{6.0 \text{ kJ}}{2 \text{ kg} \times 40.3 \text{ K}} \]Calculate this to get, \[ c_p \text{ approx } = 0.074 \text{kJ/kg·K}. \]

Key concepts

Ideal Gas Behavior

Ideal gas behavior is a model that describes how gases behave under certain conditions. An ideal gas follows the ideal gas law, which is given by: \[ PV = nRT \] where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the universal gas constant
- \( T \) is the temperature in Kelvin
In the given exercise, we assume the gas behaves ideally, which simplifies calculations. This assumption holds when the gas is at a high temperature and low pressure compared to its critical point. Ideal gas behavior allows us to directly relate pressure, volume, and temperature without accounting for intermolecular forces or the actual volume occupied by gas molecules.

Heat Transfer

Heat transfer is the process by which thermal energy moves from one object or substance to another. There are three main modes of heat transfer: conduction, convection, and radiation. In this exercise:
- We have conduction through the resistor, which heats the gas.
- We also account for the heat loss to the surroundings, which is through convection.
The heat transfer to the surroundings is calculated using the formula: \[ Q_{\text{loss}} = \text{power} \times t \] This takes into account the power rating of the heat lost to the surroundings and the duration for which the heat is lost. Understanding heat transfer allows us to understand the energy dynamics in the system and correctly balance the energy equations.

Energy Balance

An energy balance involves accounting for all the energy entering and leaving a system. In the context of the exercise, it includes the electrical work done and the heat loss to the surroundings.
Here’s how it's done:

• First, compute the electrical work done on the gas using: \[ W = V \times I \times t \]
• Then, determine the heat transferred to the surroundings: \[ Q_{\text{loss}} = \text{power} \times t \]

After finding these quantities, the net heat added to the gas is calculated: \[ Q_{\text{gas}} = W - Q_{\text{loss}} \] This energy balance ensures we accurately track the energy, leading to reliable results when computing specific heat.

Thermodynamic Properties

Thermodynamic properties are essential to understanding how systems respond to changes in their environment. In the exercise, we focus on specific heat (\(c_p\)), which is a measure of how much heat energy is needed to raise the temperature of a substance.
Specific heat at constant pressure is defined as: \[ Q = m c_p \triangle T \] where:
- \( Q \) is the heat added
- \( m \) is the mass
- \( c_p \) is the specific heat
- \( \triangle T \) is the temperature change
To determine \(c_p\), rearrange the formula: \[ c_p = \frac{Q}{m \triangle T} \] By substituting the known values, we get the specific heat. Thermodynamic properties like specific heat are crucial because they help us predict how substances will act under thermal conditions, facilitating the design and analysis of systems in engineering and science.