Solve heat transfer, gas law, engine efficiency, and entropy problems with
CalcREX's thermodynamics toolkit. 8 calculators covering Q = mcΔT, the ideal gas law, Carnot cycles,
Fourier's heat conduction, Stefan-Boltzmann radiation, and more - each with step-by-step
explanations.
From engineering design to exam preparation, these calculators
handle the core thermodynamic equations. Scroll down for in-depth topic guides with worked examples,
SVG diagrams, and links to the relevant calculators above.
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Formula:
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Result:
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How it works
Thermodynamic Constants
R (gas constant)8.314 J/(mol·K)
σ (Stefan-Boltzmann)5.67×10⁻⁸ W/(m²·K⁴)
k_B (Boltzmann)1.381×10⁻²³ J/K
c water4186
J/(kg·K)
0 °C273.15
K
1 atm101,325
Pa
Specific Heat Values
Water4,186 J/(kg·K)
Iron449 J/(kg·K)
Aluminum897 J/(kg·K)
Copper385 J/(kg·K)
Air1,005 J/(kg·K)
Glass840 J/(kg·K)
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Thermodynamics - In-Depth Guides with Worked Examples
Master heat transfer, gas behavior, engine cycles, and entropy - the physics of energy
transformation.
The Laws of Thermodynamics - The Rules Energy Must Obey
Four fundamental laws that govern all energy transformations in the
universe.
1 The Four Laws
0th Law
If A=B and B=C in thermal equilibrium, then A=C
1st Law: ΔU = Q − W
Energy is conserved - heat in = work + internal energy
2nd Law: ΔS ≥ 0
Entropy of an isolated system never decreases
3rd Law: S → 0 as T → 0 K
Perfect crystal at absolute zero has zero entropy
2 The First Law - Energy Conservation
ΔU = Q − W
ΔU = change in internal energy (J)Q = heat
added to system (+ in)W = work done BY system (+ out)
Sign convention
matters: Q is positive when heat flows INTO the system. W is positive when work
is done BY the system. Some textbooks use ΔU = Q + W (work done ON system) - always
check!
Worked Example 1
First Law - Internal Energy Change
Problem: A gas absorbs 500 J of heat and does 200 J of
work expanding against a piston. What is the change in internal energy?
Apply the First Law
ΔU = Q − W = 500 − 200
ΔU = +300 J
The gas gained 300 J of internal energy (temperature increases).
Answer: ΔU = +300 J. Of the 500 J absorbed, 200 J went
into mechanical work and 300 J raised the gas temperature.
Try the
Heat Transfer Calculator
Worked Example 2
Adiabatic Process (Q = 0)
Problem: In a rapid compression, a gas has 800 J of work
done ON it with no heat exchange. What happens to its internal energy and temperature?
Adiabatic: Q = 0, work done ON = −W
ΔU = Q − W = 0 − (−800) = +800 J
ΔU = +800 J → temperature increases
This is how diesel engines ignite fuel - compression alone heats air above the
ignition point.
Answer: Internal energy increases by 800 J, so
temperature rises. Work done ON the gas is negative W (gas didn't do the work). This is
adiabatic heating.
3 The Zeroth Law - The Foundation of Temperature
The Zeroth Law was actually formulated after the First and Second, but it is so
fundamental that it was inserted at position zero. It states: if system A is in thermal
equilibrium with system B, and system B is in thermal equilibrium with system C, then A and C
are also in thermal equilibrium.
A ⇌ B and B ⇌ C ⟹ A ⇌ C
⇌ denotes thermal equilibrium (no net heat flow)Temperature is the property that is equal when two bodies are in equilibrium
This law is the logical basis for the concept of temperature itself. Without it, we could not
claim that a thermometer reading means anything - the Zeroth Law is precisely what allows a
thermometer to work. When you put a thermometer in a cup of tea, wait for equilibrium, then read
the scale, you are applying the Zeroth Law: the thermometer and the tea are at the same
temperature, so the scale tells you the tea's temperature.
Why "zeroth"?
By the time scientists recognised this law explicitly, the First and Second were already
named and numbered. Rather than renumber everything, they called this one the Zeroth - placing
it logically before all the others.
Worked Example 3
Zeroth Law - Thermometer in Practice
Problem: A metal rod (A) is placed in contact with a
water bath (B) until no heat flows between them. The same water bath (B) is also in equilibrium
with a gas sample (C). Can we conclude the rod and the gas are at the same temperature without
putting them in direct contact?
Apply the Zeroth Law
A ⇌ B (rod and water bath in equilibrium)
B ⇌ C (water bath and gas in equilibrium)
∴ A ⇌ C - rod and gas are at the same temperature
Answer: Yes. By the Zeroth Law, A and C must be in
thermal equilibrium with each other. The water bath acted as the reference (thermometer). This
is exactly how calibrated temperature measurement works in science and engineering.
4 The Second Law - Entropy Always Increases
The Second Law is arguably the most profound law in all of physics. It introduces a direction to
time - a fundamental asymmetry. While Newton's laws work equally well forwards and backwards in
time, the Second Law does not. It tells us that in any spontaneous process in an isolated system,
entropy (disorder) never decreases.
ΔSuniverse ≥ 0
S = entropy (J/K)= 0 for reversible (ideal) processes> 0 for all real (irreversible) processes
For a reversible heat transfer at temperature T, the entropy change is:
ΔS = Q / T
Q = heat transferred (J) - positive if absorbedT = absolute temperature (K)
Clausius statement
Heat cannot spontaneously flow from cold to hot
Kelvin-Planck statement
No engine can convert heat entirely into work in a cycle
Entropy statement
ΔSuniverse ≥ 0 for every process
What entropy really means
Entropy is a measure of the number of microscopic arrangements (microstates) consistent
with the observed macrostate. Ice has few arrangements; steam has enormously many.
Melting ice spontaneously increases entropy - the universe always moves toward more
probable states.
Everyday consequences
Perfume spreading through a room, heat flowing from hot coffee to cool air, a broken egg
not reassembling - these are all entropy at work. The reverse never happens
spontaneously because the probability is astronomically small, not because it violates
any conservation law.
Arrow of time:
The Second Law is why time has a direction. If you watched a film of a cup shattering in reverse,
you would immediately know it was running backwards - because the reverse violates the Second Law.
No other law of physics would tell you this.
Worked Example 4
Entropy Change - Melting Ice
Problem: 0.5 kg of ice melts at 0°C (273.15 K). The
latent heat of fusion of water is 334,000 J/kg. Calculate the entropy change of the ice and
determine whether this process is spontaneous.
Step 1 - Heat absorbed by the ice
Q = mL = 0.5 × 334,000 = 167,000 J
Step 2 - Entropy change
ΔS = Q / T = 167,000 / 273.15
ΔS = +611.5 J/K
Answer: ΔS = +611.5 J/K. The entropy increases
(ΔS > 0), confirming the process is spontaneous in the direction of melting. The water
molecules gain far more possible arrangements as liquid than as a rigid ice crystal - exactly
what the Second Law predicts.
Worked Example 5
Second Law - Maximum Heat Engine Efficiency
Problem: A power plant operates between a boiler at
500°C and a cooling tower at 30°C. What is the maximum possible thermal efficiency, and why
can it never reach 100%?
Convert to Kelvin, apply Carnot efficiency
TH = 500 + 273.15 = 773.15 K
TC = 30 + 273.15 = 303.15 K
ηmax = 1 − TC/TH = 1 − 303.15/773.15
ηmax = 60.8%
Answer: Maximum 60.8% efficiency. The Second Law forbids
100% efficiency - some heat must always be rejected to the cold reservoir. Real power plants
achieve 35–45% due to additional irreversibilities. This is not an engineering limitation;
it is a law of nature.
5 The Third Law - The Absolute Zero Floor
The Third Law states that the entropy of a perfect crystal approaches zero as the
temperature approaches absolute zero (0 K = −273.15°C). This gives entropy an
absolute reference point - unlike energy, which is always defined relative to some arbitrary
zero.
S → 0 as T → 0 K
Valid only for a perfect, defect-free crystalReal materials retain some residual entropy at 0 K due to disorderAbsolute zero is a theoretical limit - never fully reachable
Absolute zero
0 K = −273.15°C - lowest possible temperature
Nernst theorem
ΔS → 0 for any process as T → 0 K
Unattainability
No finite sequence of steps can reach exactly 0 K
Coldest achieved
~38 picokelvin (MIT, ultracold atoms, 2021)
Why it matters practically
The Third Law allows chemists and engineers to calculate absolute entropy
values for substances - not just changes. This makes it possible to predict whether
chemical reactions will occur spontaneously using Gibbs free energy: G = H − TS. Without
absolute entropy values, this calculation would be impossible.
Superconductivity & superfluidity
As materials approach absolute zero, extraordinary quantum effects emerge. Electrical
resistance in some metals drops to exactly zero (superconductivity). Liquid helium flows
without any friction at all (superfluidity). These are direct consequences of the quantum
ground state that the Third Law describes.
Why can't we reach
absolute zero? Each cooling step removes less and less entropy, so you'd need an
infinite number of steps to reach exactly 0 K. It's like Zeno's paradox - you can halve the
remaining temperature gap forever but never close it completely. The Third Law makes this
mathematically rigorous.
Worked Example 6
Third Law - Absolute Entropy and Gibbs Free Energy
Problem: At 298 K, liquid water has an absolute molar
entropy of S° = 69.9 J/(mol·K) and an enthalpy of formation of ΔH° = −285,800 J/mol. Calculate
the Gibbs free energy of formation. Is liquid water thermodynamically stable at room
temperature?
Apply G = H − TS
G° = H° − T·S° = −285,800 − (298 × 69.9)
G° = −285,800 − 20,830
G° = −306,630 J/mol = −306.6 kJ/mol
Answer: G° = −306.6 kJ/mol. Because G° is strongly
negative, liquid water is highly stable at 298 K - it will not spontaneously decompose into
hydrogen and oxygen under normal conditions. The absolute entropy value from the Third Law was
essential to this calculation.
Heat Transfer - Conduction, Convection & Radiation
The three mechanisms by which thermal energy moves - and how to
calculate heat flow for each.
1 Three Modes of Heat Transfer
Mode
Mechanism
Formula
Conduction
Molecular vibrations through solid
Q̇ = kAΔT/d
Convection
Fluid circulation (natural or forced)
Q̇ = hAΔT
Radiation
Electromagnetic waves (no medium needed)
P = εσAT⁴
2 Sensible Heat: Q = mcΔT
Q = m × c × ΔT
Q = heat energy (J), m = mass (kg)c = specific
heat capacity (J/kg·K)ΔT = temperature change (°C or K)
Worked Example 1
Heating Water - Q = mcΔT
Problem: How much energy is needed to heat 2 kg of water
from 20°C to 100°C? (c_water = 4186 J/kg·K)
Apply Q = mcΔT
Q = 2 × 4186 × (100 − 20) = 2 × 4186 × 80
Q = 669,760 J ≈ 670 kJ
Answer: 670 kJ. A standard 2000 W kettle would take
670,000/2000 = 335 seconds ≈ 5.6 minutes. This matches real-world experience!
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Heat Transfer Calculator
Worked Example 2
Heat Conduction Through a Wall
Problem: A concrete wall is 20 cm thick, 10 m² area.
Inside is 22°C, outside is −5°C. k_concrete = 1.7 W/(m·K). What is the heat loss rate?
Fourier's Law
Q̇ = kAΔT/d = 1.7 × 10 × 27 / 0.20
Q̇ = 2,295 W ≈ 2.3 kW
Answer: 2.3 kW of heat loss through the wall alone.
Adding 10 cm of insulation (k = 0.04) would reduce this to ~100 W - a 95% reduction! This is why
insulation matters.
Try the
Heat Conduction Calculator
Worked Example 3
Stefan-Boltzmann Radiation
Problem: The Sun's surface is ~5778 K and has radius
6.96 × 10⁸ m. What is its total radiated power? (ε = 1 for blackbody)
Surface area and radiation
A = 4πr² = 4π(6.96×10⁸)² = 6.08 × 10¹⁸ m²
P = εσAT⁴ = 1 × 5.67×10⁻⁸ × 6.08×10¹⁸ × 5778⁴
P ≈ 3.85 × 10²⁶ W (385 trillion trillion watts!)
Answer: ~3.85 × 10²⁶ W - this is the Sun's luminosity.
The T⁴ dependence is key: doubling temperature increases radiation 16×.
Try the
Stefan-Boltzmann Calculator
Gas Laws - Ideal Gas, Boyle, Charles & Combined
How pressure, volume, temperature, and amount of gas relate - from
individual laws to the universal equation.
1 The Individual Gas Laws
Boyle: P₁V₁ = P₂V₂
Constant T, n
Charles: V₁/T₁ = V₂/T₂
Constant P, n
Gay-Lussac: P₁/T₁ = P₂/T₂
Constant V, n
Avogadro: V₁/n₁ = V₂/n₂
Constant T, P
Combined: P₁V₁/T₁ = P₂V₂/T₂
Constant n
2 The Ideal Gas Law
PV = nRT
P = pressure (Pa), V = volume (m³)n = moles, R
= 8.314 J/(mol·K)T = absolute temperature (K)
Worked Example 1
Ideal Gas Law - Finding Volume
Problem: What volume does 2 moles of gas occupy at 25°C
and 1 atm?
Convert and solve
T = 25 + 273.15 = 298.15 K, P = 101325 Pa
V = nRT/P = 2 × 8.314 × 298.15 / 101325
V = 0.0489 m³ = 48.9 L
Answer: 48.9 L (about 24.5 L per mole at 25°C, slightly
more than the 22.4 L/mol at STP because it's warmer).
Try the
Ideal Gas Law Calculator
Worked Example 2
Isothermal Expansion - Work Done
Problem: 1 mol of gas expands isothermally at 300 K from
10 L to 20 L. Calculate the work done by the gas.
Isothermal work formula
W = nRT × ln(V₂/V₁)
= 1 × 8.314 × 300 × ln(20/10)
= 2494.2 × 0.6931
W = 1,729 J ≈ 1.73 kJ
Answer: 1.73 kJ done by the gas. Since it's isothermal
(ΔU = 0), the gas must absorb exactly 1.73 kJ of heat - all absorbed energy goes into work.
Try the
Isothermal Work Calculator
Heat Engines - Carnot Cycle, Efficiency & Real Engines
Converting heat into work - why 100% efficiency is impossible and how
close real engines get.
1 Carnot Efficiency - The Theoretical Maximum
η_Carnot = 1 − T_C / T_H
T_C = cold reservoir temperature (K)T_H = hot
reservoir temperature (K)No real engine can exceed this efficiency
Heat engine energy flow: Heat Q_H flows from
the hot reservoir, the engine converts part to useful work W, and waste heat Q_C is rejected
to the cold reservoir. By the 1st law: Q_H = W + Q_C.
Worked Example 1
Carnot Efficiency of a Power Plant
Problem: A steam power plant operates between a boiler
at 550°C and a condenser at 30°C. (a) What is the Carnot efficiency? (b) If the plant actually
achieves 38% efficiency, what fraction of Carnot is this?
Carnot efficiency
T_H = 550 + 273 = 823 K, T_C = 30 + 273 = 303 K
η = 1 − 303/823 = 1 − 0.368
η_Carnot = 63.2%
Fraction of Carnot
38/63.2 = 60.1% of theoretical maximum
Answer: Carnot: 63.2%, actual: 38% (60% of Carnot).
Modern power plants typically achieve 35–45% efficiency. Raising T_H or lowering T_C improves
efficiency.
Try the
Carnot Efficiency Calculator
Worked Example 2
Engine Heat Rejection
Problem: A car engine absorbs 2500 J per cycle from fuel
combustion and does 750 J of work. How much heat is rejected? What is the efficiency?
Apply energy conservation
Q_C = Q_H − W = 2500 − 750
Q_C = 1,750 J rejected as waste heat
η = W/Q_H = 750/2500
η = 30%
Answer: 1,750 J waste heat, 30% efficiency. Typical
gasoline engines: 20–30%, diesel: 30–40%. The rejected heat goes into the cooling system and
exhaust - it's why radiators exist.
Entropy - Disorder, the Arrow of Time & Irreversibility
Why ice melts, heat flows from hot to cold, and broken eggs never
unbreak - it's all about entropy.
1 What Is Entropy?
Entropy (S) measures the number of microscopic arrangements (microstates) consistent with a
macroscopic state. More microstates = higher entropy = more "disorder."
ΔS = Q_rev / T
ΔS = entropy change (J/K)Q_rev = heat for a
reversible processT = absolute temperature (K)
2nd Law restated: For
any spontaneous process, ΔS_universe ≥ 0. Heat flows from hot to cold because it increases
total entropy. A refrigerator moves heat from cold to hot, but the work required increases
entropy elsewhere by even more.
Worked Example 1
Entropy Change - Melting Ice
Problem: 500 g of ice melts at 0°C. The latent heat of
fusion is 334 kJ/kg. What is the entropy change?
Calculate heat and entropy
Q = mL = 0.5 × 334,000 = 167,000 J
T = 0°C = 273.15 K
ΔS = Q/T = 167,000/273.15
ΔS = +611.3 J/K
Answer: +611.3 J/K. Positive because liquid water has
more disorder than solid ice - molecules move freely instead of being locked in a crystal
lattice.
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Entropy Calculator
Worked Example 2
Entropy of Heat Transfer Between Objects
Problem: 1000 J of heat flows from a hot object at 500 K
to a cold object at 300 K. What is the total entropy change? Is this spontaneous?
Answer: +1.33 J/K. The cold object gains more entropy
than the hot object loses - net entropy always increases for spontaneous heat flow. If heat
tried to flow from cold to hot, ΔS would be negative → impossible without external work.
Phase Changes - Latent Heat, Heating Curves & State Transitions
What happens when matter changes state - and why temperature stays
constant during melting and boiling.
1 Latent Heat
Q = mL
L_f (fusion/melting): ice = 334 kJ/kgL_v
(vaporization): water = 2260 kJ/kgTemperature stays CONSTANT during phase
change
Why L_v >> L_f:
Vaporization requires breaking ALL intermolecular bonds (molecules fly apart), while melting
only loosens the crystal structure. Boiling water needs ~6.8× more energy per kg than
melting ice.
Flat sections = phase changes
(temperature constant, all energy goes into breaking bonds). Sloped sections = temperature
changes (Q = mcΔT with different c for each phase).
Worked Example 1
Total Energy: Ice at −20°C to Steam at 120°C
Problem: Calculate the total energy to convert 1 kg of
ice at −20°C to steam at 120°C. (c_ice = 2090, c_water = 4186, c_steam = 2010 J/kg·K, L_f =
334,000, L_v = 2,260,000 J/kg)
Answer: 3.09 MJ. Vaporization alone accounts for 73% of
the total energy! This is why steam burns are so much worse than boiling water burns - steam
carries enormous latent heat.
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Heat Transfer Calculator
Worked Example 2
Thermal Equilibrium - Mixing Hot and Cold
Problem: 200 g of iron at 300°C is dropped into 500 g of
water at 20°C in an insulated container. What is the final temperature? (c_iron = 449, c_water =
4186 J/kg·K)
Answer: 31.5°C. Despite starting at 300°C, the iron
barely raises the water temperature because water's specific heat (4186) is ~9× higher than
iron's (449). Water is an exceptional heat absorber.
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