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high school physics Heat and Laws of Thermodynamics

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 and the isothermal curves appear, and 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

350 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

50 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
④W on this sketch is an assumed 0.4Q. If Q < 0 then W = 0

Heat Quantity and Specific Heat

Heat Quantity Formula
Q = mcΔT
Heat (J) = mass (kg) × specific heat (J/kg·K) × ΔT (K)
Thermal Equilibrium
Qreleased = Qabsorbed (equilibrium)
Heat lost by the hot body = heat gained by the cold body. The equilibrium formula is a box only

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?
1
Use the heat equation Q = mcΔT.
Q = mcΔT
2
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?
1
Use the first law ΔU = Q − W.
ΔU = Q − W
2
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
exam-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
1
An adiabatic process exchanges no heat, so Q = 0.
Q = 0 ⇒ ΔU = −W
2
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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Heat Engines & Efficiency
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