NDA Physics · Formula sheet

Heat and Thermodynamics formulas

12 formulas, 2 reference tables and 23 common traps for NDA Physics Heat and Thermodynamics, grouped by subtopic.

Temperature, Scales, and Thermal Expansion

Temperature and the three scales

Temperature-scale conversions

\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5} \qquad F = \frac{9}{5}C + 32 \qquad K = C + 273.15

$C$ temperature in degrees Celsius
$F$ temperature in degrees Fahrenheit
$K$ temperature in kelvin (absolute)

Thermal expansion — linear, areal, and volume coefficients

Expansion coefficients are in the ratio 1 : 2 : 3

\Delta L = L\alpha\,\Delta\theta \qquad \beta = 2\alpha \qquad \gamma = 3\alpha = \tfrac{3}{2}\beta

$\alpha$ linear expansion coefficient
$\beta$ areal (superficial) expansion coefficient
$\gamma$ volume (cubical) expansion coefficient
$\Delta\theta$ rise in temperature

Consequences of expansion — pendulums and liquid measurement

Pendulum period and apparent expansion

T = 2\pi\sqrt{\frac{L}{g}} \qquad \frac{\Delta T}{T} = \frac{1}{2}\alpha\,\Delta\theta \qquad \gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{vessel}}

$T$ time period of the pendulum
$L$ length of the pendulum rod
$\gamma_{\text{real}}$ true volume expansion of the liquid
$\gamma_{\text{apparent}}$ observed expansion (uncorrected)

Common traps

A temperature CHANGE is the same in K and °C, but a temperature VALUE is not

If the question asks 'increase of 30 K equals how many °C?', the answer is 30°C — because the step size is identical. But '300 K equals how many °C?' is

300 - 273 = 27°\text{C}

, because of the offset. Read whether it asks for a value or a change.

Never write a degree sign with Kelvin

It is '273 K' and '0 K', not '273°K'. The kelvin already IS an absolute scale, so the degree symbol is dropped by convention. An option that writes '°K' is usually the planted wrong one.

Absolute zero is −273°C, not −273 K

Absolute zero is 0 K, equivalently −273.15°C. A distractor that says 'absolute zero is −273 K' confuses the two scales — −273 K is meaningless because the Kelvin scale cannot go below 0.

Set the SCALE VARIABLES equal, not the formula sides

The condition 'two scales read the same number' means

F = C

(or

K = F

) — set those equal, THEN use a conversion to get one equation in one unknown. Do not just equate

\frac{9}{5}C+32

to something; first decide which two readings coincide.

Areal to volume is ×3/2, not ×3

Going linear to volume multiplies by 3. But going AREAL to volume multiplies by only

\frac{3}{2}

, because areal is already

2\alpha

. Decide which coefficient you were given before scaling.

Heated pendulum slows DOWN — the period goes UP

Students sometimes say 'faster' assuming heat speeds things up. Physically the rod lengthens, the period

T = 2\pi\sqrt{L/g}

rises, so the clock ticks slower and LOSES time. The effect is slight, not 'more than double'.

Heat, Specific Heat, Calorimetry, and Heat Transfer

Learn this subtopic in the notes

Specific heat, thermal capacity, and Q = mcΔT

Sensible heat (no phase change)

Q = mc\,\Delta\theta \qquad \text{Thermal capacity} = mc

$Q$ heat supplied or removed (J)
$m$ mass (kg)
$c$ specific heat capacity (J/(kg·°C))
$\Delta\theta$ change in temperature (°C or K)

Latent heat and the calorimetry mixing balance

Latent heat + the mixing balance

Q = mL \qquad \sum Q_{\text{lost}} = \sum Q_{\text{gained}}

$L$ specific latent heat (cal/g or J/kg)
$m$ mass undergoing phase change
$Q$ heat absorbed/released at constant temperature

When specific heat varies with temperature

Heat for a temperature-dependent specific heat

Q = m\int_{T_1}^{T_2}\!\big(C_0 + \alpha T\big)\,dT = m(T_2 - T_1)\left[C_0 + \tfrac{1}{2}\alpha(T_1 + T_2)\right]

$C_0$ specific heat at T = 0 (a constant)
$\alpha$ rate of change of specific heat with temperature
$T_1, T_2$ initial and final temperatures

The three modes of heat transfer — conduction, convection, radiation

Mode	How it works	Medium / key fact
Conduction	Heat passes molecule to molecule; molecules vibrate in place and pass energy to neighbours without moving from their positions	Needs a material medium; dominant in solids (especially metals)
Convection	Heated fluid becomes less dense and rises; cooler fluid sinks to replace it, setting up a circulating current that carries heat	Needs a fluid (liquid or gas) that can flow; bulk movement of matter
Radiation	Heat travels as electromagnetic (infrared) waves in a straight line	Needs NO medium; travels at the speed of light — how the Sun heats Earth NDA 2019 — 'heat waves travel in a straight line with the speed of light' is THERMAL RADIATION (not conduction or convection).

Conduction and convection both require a medium; only radiation crosses vacuum. A thermos flask defeats all three: vacuum gap stops conduction/convection, silvered walls reflect radiation.

Common traps

A body has internal energy, not 'heat'

It is loose to say a hot body 'has a lot of heat'. Strictly, heat is energy IN TRANSIT — once absorbed it becomes the body's internal energy. NDA tests the precise definition: energy transferred due to a temperature difference.

Specific heat is intrinsic; thermal capacity is not

Specific heat capacity does NOT depend on mass or shape — it is the same for 1 g or 1 tonne of the material. Thermal capacity

= mc

DOES scale with mass. The NDA repeatedly tests this distinction (statements like 'specific heat depends on mass' are false).

Watch the units — kJ vs J, kg vs g

Convert kilojoules to joules and grams to kilograms before plugging in, or the answer is off by a factor of 1000. In

Q = mc\,\Delta\theta

keep

c

in J/(kg·°C) with

m

in kg.

Don't forget the latent-heat term while ice is melting

A classic error is treating ice → water as pure

Q = mc\,\Delta\theta

. The melting itself absorbs

mL

at a flat 0°C with no temperature change. Add a separate

mL

term, or your heat budget is hundreds of calories short.

Keep masses in consistent units across both sides

In these problems one mass is in grams (ice) and the other may be given in kg (water). Convert to grams everywhere (or kg everywhere) before equating heat lost and heat gained.

The factor on $\alpha$ is one-half, not one

The planted distractor uses

[C_0 + \alpha(T_1 + T_2)]

(no one-half). The integral of

\alpha T

gives

\tfrac{\alpha}{2}T^2

, so after factoring you get

\tfrac{1}{2}\alpha(T_1 + T_2)

— the half is essential.

Only radiation crosses a vacuum

Conduction and convection BOTH need matter — conduction needs contact, convection needs a flowing fluid. Radiation alone needs no medium, which is why the Sun's heat reaches us through the vacuum of space. Any 'heat travels at the speed of light' clue means radiation.

A thermos REFLECTS radiation — it does not absorb it

The silvered walls are there to reflect infrared back, not to soak it up. An option saying the inner wall radiates and the outer absorbs is the planted wrong statement on 'which is NOT correct' questions.

Phase Change, Boiling, Evaporation, and Cooling

Learn this subtopic in the notes

Latent heat of fusion and vaporization

Latent heat

Q = mL

$Q$ heat absorbed or released at constant temperature
$m$ mass changing phase
$L$ specific latent heat (fusion or vaporization)

Boiling point depends on pressure

Boiling condition

P_{\text{vapour}} = P_{\text{atmospheric}}

$P_{\text{vapour}}$ saturated vapour pressure of the liquid
$P_{\text{atmospheric}}$ external (atmospheric) pressure

Newton's law of cooling

-\frac{d\theta}{dt} \propto (\theta - \theta_0)

$\theta$ temperature of the body
$\theta_0$ temperature of the surroundings
$\frac{d\theta}{dt}$ rate of change of temperature with time

Common traps

Latent heat is absorbed at CONSTANT temperature

The defining phrase the NDA hunts for is 'without changing the temperature'. During a phase change the thermometer reading is flat while heat is still flowing — the heat breaks molecular bonds instead of warming the substance.

Boiling is vapour pressure = atmospheric, not 'less than'

Distractors offer 'vapour pressure becomes less than' or 'greater than' atmospheric. Boiling is the exact EQUALITY: the liquid's vapour pressure rises to meet the external pressure. Pick the 'equal to' option.

Evaporation happens at ALL temperatures; boiling does not

A common confusion is that a liquid only evaporates when heated. Evaporation occurs at any temperature (wet clothes dry in the shade). Boiling is the one that needs the specific boiling-point temperature.

Newton's law does NOT apply to phase changes or furnace heat

The law needs the body's temperature to actually be changing and the temperature difference to be small. Melting ice and boiling water hold a constant temperature; a furnace-hot rod has a huge difference dominated by radiation. Only gentle convective cooling qualifies.

Gas Laws and the Laws of Thermodynamics

Learn this subtopic in the notes

The ideal gas law

Ideal gas law and the combined gas law

PV = nRT \qquad \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

$P$ pressure
$V$ volume
$n$ number of moles (or molecules)
$R$ universal gas constant
$T$ absolute temperature (K)

First law of thermodynamics

\Delta U = Q - W

$\Delta U$ change in internal energy
$Q$ heat supplied to the system
$W$ work done BY the system

Named processes — isothermal, adiabatic, isochoric, isobaric

Identify a process by substituting PV = nRT

P = kT \;\Rightarrow\; V = \frac{nR}{k} = \text{const} \;\Rightarrow\; \text{isochoric},\; C = C_V

$k$ the constant in the given process equation
$C_V$ molar heat capacity at constant volume
$C_P$ molar heat capacity at constant pressure

The second law and a process summary table

Process / law	What is held / stated	Key consequence
Isothermal	Temperature constant	$\Delta U = 0$ ; $PV = \text{const}$ ; all heat becomes work
Adiabatic	No heat exchanged (Q = 0)	Insulated; temperature still changes (compression heats the gas)
Isochoric	Volume constant (W = 0)	$\Delta U = Q$ ; molar heat capacity $C_V$ ; $P \propto T$
Isobaric	Pressure constant	Molar heat capacity $C_P$ (and $C_P > C_V$ ); $V \propto T$
Second law	Heat won't flow cold → hot unaided	External work needed to move heat uphill (refrigerator); sets the direction of natural processes NDA 2017 — 'heat cannot flow by itself from a lower to a higher temperature' is the SECOND law of thermodynamics.

The first law is energy bookkeeping (ΔU = Q − W); the second law sets the one-way direction of heat flow.

Common traps

Temperature in the gas law is ALWAYS in kelvin

Using Celsius in

PV = nRT

or in

P \propto T

gives wrong ratios. Convert to kelvin first. 'Pressure doubles when temperature doubles' is only true on the absolute scale.

Internal energy of an ideal gas depends only on temperature

In an ISOTHERMAL process (constant T) the internal energy does not change at all (

\Delta U = 0

), so any heat absorbed is entirely converted to work. Don't assume absorbing heat always raises internal energy.

Don't guess the process — substitute PV = nRT

A process given as

P = kT

PV^2 = \text{const}

is not one of the four standard names on sight. Substitute the ideal gas law to see which variable is actually constant, then read off the heat capacity or T-V relation.

Adiabatic means no HEAT exchange, not no temperature change

An adiabatic process has

Q = 0

but the temperature usually DOES change (an adiabatic compression heats a gas). 'No heat exchange' is the definition; 'constant temperature' is isothermal — a different process.

First law = energy; second law = direction

The first law (ΔU = Q − W) is conservation of energy and is direction-blind. The second law adds the arrow: heat flows hot → cold spontaneously, never the reverse without work. Statements about 'cannot flow by itself' point to the SECOND law.

Heat and Thermodynamics formulas

Temperature, Scales, and Thermal Expansion

Temperature and the three scales

Thermal expansion — linear, areal, and volume coefficients

Consequences of expansion — pendulums and liquid measurement

Common traps

Heat, Specific Heat, Calorimetry, and Heat Transfer

Specific heat, thermal capacity, and Q = mcΔT

Latent heat and the calorimetry mixing balance

When specific heat varies with temperature

The three modes of heat transfer — conduction, convection, radiation

Common traps

Phase Change, Boiling, Evaporation, and Cooling

Latent heat of fusion and vaporization

Boiling point depends on pressure

Common traps

Gas Laws and the Laws of Thermodynamics

The ideal gas law

Named processes — isothermal, adiabatic, isochoric, isobaric

The second law and a process summary table

Common traps

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