Question

A gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of \(1000 \mathrm{~W}\). Heat transfer between the gas and the surroundings occurs at a rate of \(\dot{Q}=-50 t\), where \(\dot{Q}\) is in watts, and \(t\) is time, in min. (a) Plot the time rate of change of energy of the gas for \(0 \leq t \leq 20 \mathrm{~min}\), in watts. (b) Determine the net change in energy of the gas after 20 min, in kJ. (c) If electricity is valued at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), what is the cost of the electrical input to the resistor for \(20 \mathrm{~min}\) of operation?

Short answer

a) The rate of energy change decreases linearly from 1000 W to 0 W. b) Net change is 600 kJ. c) The cost is \$2.67.

Step-by-step solution

Understand the Given Information

The electric resistor transfers energy to the gas at a constant rate of 1000 W. Heat transfer between the gas and surroundings occurs at a rate of \(\dot{Q} = -50t\), where \(\t\) is in minutes.

Determine Time Rate of Change of Energy (Part a)

The rate of change of energy of the gas (\(\frac{dE}{dt}\)) is the difference between the rate of energy addition by the resistor and the rate of heat transfer. Therefore, \[ \frac{dE}{dt} = 1000 - 50t \] Plot \(\frac{dE}{dt} \) versus \t\ for 20 minutes.

Create the Plot (Part a)

Use a plotting tool or hand-draw the graph. \(\frac{dE}{dt}\) starts at 1000 W when \t = 0\ and decreases linearly to \(-50 \times 20 + 1000 = 0\) W when \t = 20\.

Calculate Net Change in Energy (Part b)

Integrate \(\frac{dE}{dt}\) over \t\ from 0 to 20 minutes to find the net change in energy. \[ \text{Net Change in Energy} = \int_0^20 (1000 - 50t) dt \] Solve the integral: \[ \int (1000 - 50t) dt = 1000t - 25t^2 \] Evaluate from 0 to 20: \[ (1000 \times 20 - 25 \times 20^2) - (1000 \times 0 - 25 \times 0^2) = 20000 - 10000 = 10000 \] Therefore, the net change in energy is 10000 W-min. Convert to kJ: \[ 10000 \times \frac{60}{1000} = 600 \] kJ.

Calculate the Cost of Electrical Input (Part c)

First, determine the total energy input by the resistor in kWh: \[ 1000 \times \frac{20}{60} = \frac{1000 \times 20}{60} = \frac{20000}{60} \] which is approximately \(\frac{200}{6} \) kWh. Multiply the energy input by the cost of electricity: \[ \frac{200}{6} \times 0.08 = \frac{16}{6} \approx 2.67 \] dollars.

Key concepts

energy transfer

Energy transfer is a fundamental concept in thermodynamics. It describes how energy moves from one place to another.
In this exercise, energy transfer happens in two main ways:

• The electric resistor transfers energy to the gas at a constant rate of 1000 W.
• Heat transfer occurs between the gas and its surroundings, given by the formula \(\frac{dQ}{dt} = -50t\). This negative value indicates that heat is leaving the gas.

Understanding how energy is transferred is key to solving problems like this one. It helps you determine how and where energy is moving, which is essential for accurate calculations.

heat transfer

Heat transfer refers to the movement of thermal energy from one object or substance to another. This can happen through conduction, convection, or radiation. In our exercise, the heat transfer rate is given by \(\frac{dQ}{dt} = -50t\) W, where \( t \) is in minutes.
This equation tells us two important things:

• Heat leaves the gas (because of the negative sign).
• The rate at which heat is leaving increases with time (proportional to \( t \)).

This dynamic plays a crucial role in determining the overall energy balance in the system.

rate of change of energy

The rate of change of energy (\frac{dE}{dt}) is the sum of energy added by the resistor and the energy lost due to heat transfer. In our problem, this is formulated as: \[(\frac{dE}{dt}) = 1000 - 50t \]
This means:

• At \(t = 0\), the rate of change of energy starts at 1000 W.
• As time goes on, the rate decreases linearly, reaching 0 W at \(t= 20\) minutes.

This concept helps us understand how quickly or slowly the energy in the gas is changing over time, which is crucial for accurate calculations.

integration in thermodynamics

Integration in thermodynamics is used to find the total change in a quantity over a period of time. In this exercise, we integrated the rate of change of energy to find the net change in energy over 20 minutes.
The integral we solved was:

• \[\text{Net Change in Energy} = \displaystyle\int_0^{20}(1000 - 50t) dt = 1000t - 25t^2 \]

Evaluating this from 0 to 20 minutes, we found the net change in energy to be 600 kJ.
Integration helps us move from rates of change to total amounts, which is key for understanding the overall energy changes in a thermodynamic system.