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PhysicsThermodynamics
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be :
Options
1
75 W
2
100 W
3
25 W
4
125 W
Correct Answer
Option 3 : 25 W
Step-by-Step Solution
1

First Law of Thermodynamics (rate form):

\(\dfrac{dU}{dt} = \dfrac{dQ}{dt} - \dfrac{dW}{dt}\)

Where: dQ/dt = rate of heat supplied = 100 W (into system)

dW/dt = rate of work done BY system = 75 J/s

2

Calculate rate of internal energy increase:

\(\dfrac{dU}{dt} = 100 - 75 = \mathbf{25 \text{ W}}\) ✓

Physical interpretation:

Of the 100 J of heat entering the system every second, 75 J/s leaves as work done by the system. The remaining 25 J/s stays in the system as increased internal energy (temperature rise).

Theory: First Law of Thermodynamics
1. Statement of First Law

The First Law of Thermodynamics is a statement of energy conservation for thermodynamic systems. It states that the change in internal energy of a system equals the heat added to the system minus the work done by the system:

\(\Delta U = Q - W\) (Physics convention)

\(\Delta U = Q + W\) (Chemistry convention: W = work done ON system)

In NEET (and this problem), the physics convention is used: Q is heat added to system (positive if absorbed), W is work done by system (positive if system expands). The first law essentially says energy cannot be created or destroyed — only converted between forms.

2. Rate Form of First Law

Dividing by time dt: dU/dt = dQ/dt − dW/dt. This gives the power form — all terms are in watts (J/s). This is exactly what this problem uses: heat supply rate (100 W) minus work rate (75 W) = internal energy rate (25 W).

3. Special Thermodynamic Processes

📌 Isothermal: T = const → ΔU = 0 → Q = W (ideal gas)

📌 Adiabatic: Q = 0 → ΔU = −W (work done at expense of internal energy)

📌 Isochoric: V = const → W = 0 → ΔU = Q (all heat goes to internal energy)

📌 Isobaric: P = const → W = PΔV, ΔU = Q − PΔV

4. Internal Energy

Internal energy U is the sum of kinetic and potential energies of all molecules in the system. For an ideal gas: U = nCᵥT (depends only on temperature). For real gases, U also depends on volume (due to intermolecular forces). When U increases, temperature rises (for ideal gas). Internal energy is a state function — depends only on current state, not on how it was reached.

5. Work Done by Gas

W = ∫P dV (general)

📌 Isobaric: W = PΔV = P(V₂−V₁)

📌 Isothermal: W = nRT ln(V₂/V₁)

📌 Adiabatic: W = (P₁V₁−P₂V₂)/(γ−1) = nCᵥ(T₁−T₂)

📌 Isochoric: W = 0 (no volume change)

6. Molar Heat Capacities

Cᵥ = heat capacity at constant volume = ΔU/ΔT per mole. Cₚ = heat capacity at constant pressure > Cᵥ (extra energy for PΔV work). Relation: Cₚ − Cᵥ = R (Mayer's relation). Ratio γ = Cₚ/Cᵥ: monoatomic gas (5/3 ≈ 1.67), diatomic (7/5 = 1.4), triatomic (4/3 ≈ 1.33).

⚠️ Physics convention: ΔU = Q − W (W = work by system)

⚠️ Chemistry convention: ΔU = Q + W (W = work on system)

⚠️ NEET uses physics convention — be consistent throughout!

⚠️ Heat into system: Q positive. Work by system: W positive.

Frequently Asked Questions
1. How is the First Law applied in rate form?
dU/dt = dQ/dt − dW/dt. Heat supply rate = 100 W (in), work rate = 75 W (out), so dU/dt = 100 − 75 = 25 W. The 25 W increases the internal energy of the system every second.
2. What happens to the 75 J/s of work done?
The 75 J/s work done BY the system goes into the surroundings (e.g., pushing a piston, rotating a shaft). It leaves the system. The remaining 25 J/s stays in the system as increased internal energy.
3. What is the sign convention for Q in NEET?
Q is positive when heat flows INTO the system (system absorbs heat). Q is negative when heat flows OUT of the system (system releases heat). In this problem, the heater adds 100 W to the system, so Q = +100 J/s.
4. What is an adiabatic process?
Q = 0 — no heat exchange with surroundings. By First Law: ΔU = −W. If gas expands adiabatically (does positive work), internal energy decreases and temperature falls. Adiabatic processes are fast — no time for heat exchange. Example: sound propagation in air.
5. Why is Cₚ > Cᵥ?
At constant pressure, heat supplied goes into both increasing internal energy AND doing PΔV work of expansion. At constant volume, all heat goes only into internal energy. So more heat is needed at constant pressure for same temperature rise: Cₚ > Cᵥ. Difference = R (gas constant) per mole.
6. What is Mayer's relation?
Cₚ − Cᵥ = R, where R = 8.314 J/mol·K is the universal gas constant. For monoatomic ideal gas: Cᵥ = (3/2)R, Cₚ = (5/2)R. For diatomic: Cᵥ = (5/2)R, Cₚ = (7/2)R. γ = Cₚ/Cᵥ.
7. Can internal energy decrease?
Yes. If ΔU = Q − W is negative, internal energy decreases. This happens when work done by system exceeds heat added: e.g., adiabatic expansion (Q=0, W>0 → ΔU<0, temperature drops). Adiabatic cooling is why temperature drops at high altitudes.
8. What is the Second Law of Thermodynamics?
Heat cannot spontaneously flow from cold to hot body (Clausius). No heat engine can have 100% efficiency (Kelvin-Planck). Entropy of an isolated system never decreases. Efficiency of Carnot engine η = 1 − T_cold/T_hot (maximum possible for any heat engine between those temperatures).
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