Heat and Laws of Thermodynamics

Heat is energy transferred because of a temperature difference, so it flows when hot and cold bodies touch. Q = mcΔT multiplies mass, specific heat, and temperature change, and the first law ΔU = Q − W is energy conservation for a gas. On a P-V diagram the area under the path is the work done by the gas. Change temperature and heat supplied here to watch isothermal curves and how energy is shared.

What is Heat?

🔥 Intuition of Heat

①Heat flows when hot and cold objects touch

②Heat (Q) is energy transferred due to temperature difference

③Heat amount Q = mcΔT — mass × specific heat × temperature change

P-V Diagram

Temperature T₂350 K

💡 Reading a P-V Diagram

①Isothermal: PV = nRT = const → hyperbola

②Higher temperature → curve shifts up/right

③Area under P-V graph = work done by gas (W)

First Law of Thermodynamics

Heat Supplied Q50 J

First Law (Energy Conservation)

ΔU = Q - W

Change in internal energy = absorbed heat − work done by gas

⚡ Meaning of the First Law

①Energy is neither created nor destroyed (conserved!)

②Heat (Q) into gas splits into ΔU + work (W)

③Q > 0: absorbed, Q < 0: released / W > 0: gas does work

Heat Quantity and Specific Heat

Heat Quantity Formula

Q = mcΔT

Heat (J) = mass (kg) × specific heat (J/kg·K) × ΔT (K)

Thermal Equilibrium

Q_released = Q_absorbed (equilibrium)

Heat lost by the hot body = heat gained by the cold body

Worked Examples

Example 1

How much heat is needed to raise the temperature of a 2 kg object with specific heat 0.5 kJ/(kg·K) by 20 K?

Use the heat equation Q = mcΔT.

Q = mcΔT

Substitute m = 2 kg, c = 0.5 kJ/(kg·K), ΔT = 20 K.

Q = 2 × 0.5 × 20 = 20 kJ

▸ 20 kJ

A larger specific heat c needs more heat for the same temperature change. Keep the units (kJ/kg·K) consistent.

Example 2

A gas absorbs 80 J of heat and does 30 J of work on its surroundings. What is the change in internal energy?

Use the first law ΔU = Q − W.

ΔU = Q − W

Substitute Q = +80 J (absorbed), W = +30 J (work by gas).

ΔU = 80 − 30 = 50 J

▸ 50 J (increase)

In the first law, W is the work done by the gas. The heat left over after doing work raises the internal energy.

Summary

Heat Amount

Q = mcΔT

First Law

ΔU = Q - W

CSAT-style

A gas in an insulated container does 120 J of work on its surroundings. What is the change in its internal energy?

①+120 J

②+60 J

③0 J

④−60 J

⑤−120 J

▸ ⑤ −120 J

An adiabatic process exchanges no heat, so Q = 0.

Q = 0 ⇒ ΔU = −W

Substitute W = +120 J (work done by the gas).

ΔU = −120 J

🎯 Exam Points

①In Q = mcΔT, larger c → smaller ΔT

②First law: ΔU = Q - W (energy conservation)

③Isothermal: ΔT=0 → ΔU=0 → Q=W

④Adiabatic: Q=0 → ΔU=-W (doing work cools the gas)

⑤Heat flows spontaneously only from hot to cold (Second law)

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Conservation of Mechanical Energy

Heat Engines & Efficiency

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