Heat, Specific Heat & the First Law

You’ll be able to

Distinguish heat, temperature, and internal energy
Use Q = mcΔT to calculate heat transfer and specific heat
Apply the first law of thermodynamics with correct signs for heat and work

Heat is energy in transit

Heat (Q) is thermal energy that flows from a hotter object to a colder one because of their temperature difference. It is not something an object “contains” — that stored quantity is internal energy (U), the total kinetic energy of all the molecules. Temperature, meanwhile, is the average molecular kinetic energy. A bathtub of warm water holds far more internal energy than a spark, even though the spark is at a much higher temperature.

Heat and temperature change

Q = m · c · ΔT

c is the specific heat (J·(kg·K)⁻¹): the heat needed to raise 1 kg by 1 K. ΔT = T_final − T_initial, so Q is positive when the object is heated.

Why water resists temperature change

Specific heat measures thermal “inertia.” Water’s is unusually large — about 4186 J·(kg·K)⁻¹, roughly 4–5 times that of most metals — so the same heat input barely warms water while it sends a metal’s temperature soaring. This is why coastal climates are mild and why a metal spoon in soup burns your hand before the broth does. For a fixed heat Q, a smaller specific heat means a larger temperature change.

Worked example

How much heat is required to raise the temperature of 0.50 kg of water by 20 °C? (c = 4186 J·(kg·K)⁻¹)

1.A change of 20 °C equals a change of 20 K, so ΔT = 20.
2.Apply Q = mcΔT = (0.50 kg)(4186 J·(kg·K)⁻¹)(20 K).
3.Multiply: 0.50 × 4186 × 20 = 41 860 J.

Answer: ≈ 4.19 × 10⁴ J (about 41.9 kJ)

Tip

A temperature difference is identical in Celsius and kelvin (a rise of 20 °C is a rise of 20 K), so ΔT never needs converting in Q = mcΔT. Only convert when a formula uses an absolute temperature, not a change.

Checkpoint

How much heat is needed to raise the temperature of 0.20 kg of aluminum by 30 °C? (c = 900 J·(kg·K)⁻¹)

First law of thermodynamics

ΔU = Q − W

Q is heat added TO the gas; W is work done BY the gas. Adding heat raises internal energy; letting the gas do work (expand) lowers it.

The first law is energy conservation

The first law says a gas’s internal energy changes only by two routes: heat crossing the boundary and work crossing it. Sign conventions are everything. Q > 0 when heat flows in; W > 0 when the gas expands and pushes on its surroundings. If the surroundings instead compress the gas, they do work on it — the gas does negative work, so W < 0 and that energy stays inside, raising U.

Checkpoint

A gas is compressed by an external force that does 300 J of work on it, while no heat is exchanged (Q = 0). What is the change in the gas’s internal energy?

On the exam

Read the wording carefully: “work done ON the gas” is the opposite sign of the W in ΔU = Q − W (which is work done BY the gas). Compression → gas does negative work → U tends to rise; expansion → gas does positive work → U tends to fall.

Answer the 2 checkpoints as you read.