Why Hot Chai Cools Quickly at First and Then Slowly

Stir hot dal with a steel spoon and the handle soon warms. Put a pot on the stove and the water churns. Sit near a bonfire and your face glows warm even though the air between is cool.

These are the three ways heat travels — and each follows its own rule.

This part covers the three modes, thermal conductivity, blackbody radiation with Wien's and Stefan's laws, and Newton's law of cooling.

Conduction passes heat from particle to particle without the material moving, convection carries heat by the bulk flow of a fluid, and radiation carries heat as electromagnetic waves that need no medium at all.

- Conduction — mainly in solids: a steel spoon in hot tea, the base of a kadhai
- Convection — in liquids and gases: water circulating in a pot, sea and land breezes
- Radiation — through empty space: sunlight, warmth from a heater

Worked example — radiation needs no medium. Sunlight crosses about $1.5\times 10^{11}$ m of nearly empty space at $3\times 10^8$ m/s:

$$t=\frac{1.5\times 10^{11}}{3\times 10^8}=500\text{ s}\approx 8.3\text{ minutes}$$

No conduction or convection could carry heat across a vacuum.

An everyday example. Frying pakoras in a kadhai uses all three: the metal conducts heat from the flame, the oil carries it round by convection, and you feel radiation on your hands above the pan.

The substance. Natural convection depends on gravity — warm fluid rises because it is less dense, so without gravity only forced convection, such as a fan, works.

**In the steady state, the rate of heat flow through a slab is $H=\frac{KA(T_1-T_2)}{L}$, where $K$ is the thermal conductivity; for slabs in series, the same $H$ flows through each layer and their thermal resistances $R=\frac{L}{KA}$ add.

Worked example 1 — a window.** A glass pane has area $2$ m${}^2$, thickness $5$ mm and $K=0.8$ W/m K, with $25$ °C inside and $5$ °C outside.

$$H=\frac{0.8\times 2\times 20}{0.005}=6400\text{ W}$$

Worked example 2 — a composite wall. A $1$ m${}^2$ wall has $0.2$ m of brick ($K=0.6$) and $0.05$ m of insulation ($K=0.05$), with $40$ °C on the brick side and $10$ °C on the other.

$$R_1=\frac{0.2}{0.6}\approx 0.333\text{ K/W}{R}_2=\frac{0.05}{0.05}=1\text{ K/W}$$

$$H=\frac{30}{1.333}=22.5\text{ W}$$

The temperature drop across the brick is $22.5\times 0.333=7.5$ K, so the **interface is at $32.5$ °C.

An everyday example. A thermocol box keeps kulfi frozen because thermocol's very low conductivity gives a large thermal resistance.

The substance. Steady state means temperatures do not change with time** — not that the temperature is the same everywhere.

**A blackbody absorbs all radiation falling on it and emits a continuous spectrum whose peak wavelength obeys Wien's law, ${\lambda }_{m}T=b$ with $b\approx 2.9\times 10^{-3}$ m K, and whose total power obeys Stefan's law, $P=e\sigma A{T}^4$ with $\sigma =5.67\times 10^{-8}$ W/m${}^2$ K${}^4$.

Features: the spectrum is continuous; as temperature rises, emission rises at every wavelength and the peak shifts to shorter wavelengths.** A body in surroundings at $T_0$ loses energy at the net rate $e\sigma A(T^4-T_0^4)$.

Worked example 1 — the Sun. Its spectrum peaks near $500$ nm:

$$T=\frac{2.9\times 10^{-3}}{5\times 10^{-7}}=5800\text{ K}$$

Worked example 2 — the human body. At $310$ K:

$${\lambda }_m=\frac{2.9\times 10^{-3}}{310}\approx 9.4\times 10^{-6}\text{ m}$$

That is in the infrared, which is why thermal cameras can see people in the dark.

Worked example 3 — Stefan's law. A black sphere of radius $0.1$ m at $600$ K:

$$P=5.67\times 10^{-8}\times 4\pi (0.1{)}^2\times 600^4\approx 923\text{ W}$$

Doubling the kelvin temperature multiplies the power by $2^4=16$.

An everyday example. An iron rod in a blacksmith's forge glows dull red, then orange, then yellow-white as it heats — the peak moving to shorter wavelengths.

The substance. Every body above absolute zero radiates; it stays warm only if it absorbs as much as it emits.

**For small temperature differences, the rate of cooling of a body is proportional to how much hotter it is than its surroundings, $\frac{dT}{dt}=-k(T-T_s)$, so it cools quickly at first and more slowly as it approaches room temperature.**

Integrating gives an exponential curve, $T-T_s=(T_0-T_s)e^{-kt}$. For quick estimates, use the average temperature over an interval:

$$\frac{T_1-T_2}{t}=k\left(\frac{T_1+T_2}{2}-T_s\right)$$

Worked example. A body cools from $80$ °C to $60$ °C in $5$ min in a room at $20$ °C. How long does it take to cool from $60$ °C to $40$ °C?

**Find $k$.**

$$\frac{20}{5}=k(70-20)\Rightarrow k=0.08\text{ per min}$$

Second interval.

$$\frac{20}{t}=0.08(50-20)=2.4\Rightarrow t\approx 8.3\text{ min}$$

The **same $20$ degree drop takes longer because the body is now closer to room temperature.

An everyday example. A cup of chai is too hot to sip for the first few minutes, then seems to stay lukewarm for a long time.

The substance. A graph of $\ln (T-T_s)$ against time is a straight line** with slope $-k$.