Question:medium
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 W, then the rate at which internal energy increases will be: ____.
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 W, then the rate at which internal energy increases will be: ____.
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Internal energy is like a bank account. If you put in \$100 (heat) but spend \$75 (work), your balance (internal energy) only increases by \$25.
Updated On: May 28, 2026
- 125 W
- 75 W
- 100 W
- 25 W
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Topic:
This problem is an application of the "First Law of Thermodynamics." The First Law is essentially a statement of the conservation of energy. It tells us that energy added to a system as heat must either increase the internal energy of the system or be used by the system to do work on its surroundings.
Step 2: Key Formulas and Approach:
The First Law is usually written as: \[ \Delta Q = \Delta U + \Delta W \] When dealing with rates (energy per unit time), we take the derivative with respect to time ($t$): \[ \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \] Where $dQ/dt$ is the power input as heat, $dW/dt$ is the mechanical power output, and $dU/dt$ is the rate of change of internal energy.
Step 3: Detailed Explanation:
Identify given values: Heat supply rate ($dQ/dt$) = $100 \text{ W}$ (positive because heat is entering). Work performance rate ($dW/dt$) = $75 \text{ W}$ (positive because work is being done by the system).
Setup the equation: We want to find the rate of internal energy increase ($dU/dt$). \[ 100 \text{ W} = \frac{dU}{dt} + 75 \text{ W} \]
Solve for the unknown: Subtract the work rate from the heat rate. \[ \frac{dU}{dt} = 100 \text{ W} - 75 \text{ W} = 25 \text{ W} \]
This means that out of the 100 Joules of energy entering every second, 75 Joules are converted into external work, and the remaining 25 Joules are stored as internal kinetic or potential energy of the particles in the system.
Step 4: Final Answer:
The rate at which internal energy increases is 25 W.
This problem is an application of the "First Law of Thermodynamics." The First Law is essentially a statement of the conservation of energy. It tells us that energy added to a system as heat must either increase the internal energy of the system or be used by the system to do work on its surroundings.
Step 2: Key Formulas and Approach:
The First Law is usually written as: \[ \Delta Q = \Delta U + \Delta W \] When dealing with rates (energy per unit time), we take the derivative with respect to time ($t$): \[ \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \] Where $dQ/dt$ is the power input as heat, $dW/dt$ is the mechanical power output, and $dU/dt$ is the rate of change of internal energy.
Step 3: Detailed Explanation:
Identify given values: Heat supply rate ($dQ/dt$) = $100 \text{ W}$ (positive because heat is entering). Work performance rate ($dW/dt$) = $75 \text{ W}$ (positive because work is being done by the system).
Setup the equation: We want to find the rate of internal energy increase ($dU/dt$). \[ 100 \text{ W} = \frac{dU}{dt} + 75 \text{ W} \]
Solve for the unknown: Subtract the work rate from the heat rate. \[ \frac{dU}{dt} = 100 \text{ W} - 75 \text{ W} = 25 \text{ W} \]
This means that out of the 100 Joules of energy entering every second, 75 Joules are converted into external work, and the remaining 25 Joules are stored as internal kinetic or potential energy of the particles in the system.
Step 4: Final Answer:
The rate at which internal energy increases is 25 W.
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