seegongsik
Grade 10-11 (age 15-17)

high school physics Heat and Laws of Thermodynamics

Heat is energy that moves because of a temperature difference, so when a hot object touches a cold one, heat flows from the hot side to the cold side. The heat an object takes in is found from Q = mcΔT, the product of mass, specific heat and temperature change, and the heat a gas takes in goes into raising its internal energy or doing work on the outside, so ΔU = Q − W. On a P-V graph, the area under the curve is the work the gas does as it expands. Changing the temperature moves the isotherms, and changing the heat Q and the work W shows how the change in the gas's internal energy is decided.

What Heat Is

🔥 Energy That Flows Down a Temperature Difference
①When a hot object touches a cold one, heat moves from the hot side to the cold side until the two temperatures are equal
②Heat Q is energy transferred because of a temperature difference, measured in J
③The heat needed to raise an object's temperature by ΔT is Q = mcΔT, proportional to the mass m and the specific heat c

The P-V Graph and Work

350 K
💡 Reading the Graph
①At constant temperature PV is constant, so each curve is a hyperbola
②The higher the temperature, the higher the pressure at the same volume, so the curve lies farther from the origin
③The shaded area under the curve is the work W the gas does on the outside as it expands, and raising T₂ makes this area larger

The First Law of Thermodynamics

50 J
20 J
First Law (Energy Conservation)
ΔU = Q − W
Change in internal energy = heat into the gas − work done by the gas
⚡ Where the Heat Goes
①Energy is neither created nor destroyed, so the heat Q into the gas splits into the internal energy increase ΔU and the work W done on the outside
②Q positive means heat is taken in and negative means it is given out; W positive means the gas expands and does work, and negative means it is compressed and has work done on it
③When Q and W are equal, ΔU = 0 and the temperature stays the same; when W is negative, the work received adds on, so the internal energy rises by more than Q

Heat and Specific Heat

Heat
Q = mcΔT
Heat (J) = mass (kg) × specific heat (J/(kg·K)) × temperature change (K)
Thermal Equilibrium
Qreleased = Qabsorbed (equilibrium)
If no heat leaks out, the heat lost by the hotter object equals the heat gained by the cooler one, and heat stops flowing once the two temperatures are equal

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 formula Q = mcΔT.
Q = mcΔT
2
Substitute m = 2 kg, c = 0.5 kJ/(kg·K) and ΔT = 20 K.
Q = 2 × 0.5 × 20 = 20 kJ
▸ 20 kJ
The specific heat is given in kJ, so the answer is in kJ too. Raising the same object by 40 K would take twice as much, 40 kJ.
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 (heat taken in) and W = +30 J (work done by the gas).
ΔU = 80 − 30 = 50 J
▸ 50 J (increase)
Of the 80 J of heat taken in, 30 J leaves as work, and the remaining 50 J raises the internal energy.

Summary

Heat
Q = mcΔT
First Law
ΔU = Q − W
CSAT-style problem, adapted
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
In an adiabatic process no heat goes in or out, 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, the larger the specific heat c, the smaller the temperature change for the same heat
②The first law ΔU = Q − W expresses energy conservation
③In an isothermal process ΔT = 0, so ΔU = 0 and Q = W
④In an adiabatic process Q = 0, so ΔU = −W, and a gas that expands and does work cools down
⑤On its own, heat flows only from hot to cold (the second law of thermodynamics)

Last updated

Was this helpful? Support seegongsik