Question
3. A 2.0 kg quantity of water is held at constant volume in a pressure cooker and heated by a range element. The system's internal energy increases by 8.0 × 103 J. However, the pressure cooker is not well insulated, and as a result, 2.0 x 103J of energy is transferred to the surrounding air. How much energy is transferred from the range element to the pressure cooker as heat?
Answers
GPT-4o mini
To determine the amount of energy transferred from the range element to the pressure cooker as heat, we can apply the first law of thermodynamics, which is expressed as:
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.
In this scenario, since the volume is held constant, there is no work done by the system (\(W = 0\)). Therefore, the equation simplifies to:
\[
\Delta U = Q
\]
We know the following from the problem statement:
- The change in internal energy, \(\Delta U = 8.0 \times 10^3 \, \text{J}\),
- The energy lost to the surroundings is \(2.0 \times 10^3 \, \text{J}\).
Since energy is conserved, we can express the heat added to the system, \(Q\), as the sum of the internal energy change and the energy lost to the surroundings:
\[
Q = \Delta U + \text{Energy lost to surroundings}
\]
Substituting in the known values:
\[
Q = (8.0 \times 10^3 \, \text{J}) + (2.0 \times 10^3 \, \text{J})
\]
\[
Q = 10.0 \times 10^3 \, \text{J}
\]
Thus, the amount of energy transferred from the range element to the pressure cooker as heat is:
\[
Q = 10,000 \, \text{J} \text{ or } 1.0 \times 10^4 \, \text{J}
\]
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.
In this scenario, since the volume is held constant, there is no work done by the system (\(W = 0\)). Therefore, the equation simplifies to:
\[
\Delta U = Q
\]
We know the following from the problem statement:
- The change in internal energy, \(\Delta U = 8.0 \times 10^3 \, \text{J}\),
- The energy lost to the surroundings is \(2.0 \times 10^3 \, \text{J}\).
Since energy is conserved, we can express the heat added to the system, \(Q\), as the sum of the internal energy change and the energy lost to the surroundings:
\[
Q = \Delta U + \text{Energy lost to surroundings}
\]
Substituting in the known values:
\[
Q = (8.0 \times 10^3 \, \text{J}) + (2.0 \times 10^3 \, \text{J})
\]
\[
Q = 10.0 \times 10^3 \, \text{J}
\]
Thus, the amount of energy transferred from the range element to the pressure cooker as heat is:
\[
Q = 10,000 \, \text{J} \text{ or } 1.0 \times 10^4 \, \text{J}
\]