Heat dissipation and cooling solutions: Exercises — Hardware deep dive

This page is a self-test ladder for Heat dissipation and cooling solutions. Every problem hides its answer inside a collapsible Solution box — try first, then reveal. Difficulty climbs from recognising a formula to synthesising a whole cooling system.

Definition The symbols you will reuse everywhere

$T_{amb}$ = ambient temperature: the temperature of the surrounding air (or room) that the cooler ultimately dumps heat into. It's the baseline everything is measured above — a chip can never get cooler than its ambient.
$T_{j}$ = junction temperature: the temperature of the hottest active silicon inside the chip (the "junction" is where transistors live). It's the number we actually care about protecting — it equals ambient plus the total temperature rise: $T_{j} = T_{amb} + Δ T$ . When $T_{j}$ crosses the chip's spec, Thermal throttling triggers.
$k$ = thermal conductivity: a material property telling you how readily heat crawls through a solid. Units W/m·K (watts of heat per metre of thickness per kelvin of temperature difference). Copper $k = 400$ , aluminium $k = 205$ , air $k \approx 0.025$ — metals are ~10,000× better than air.
$h$ = heat-transfer coefficient: how readily a moving fluid (air, water) strips heat off a surface. Units W/m²·K (watts per square metre of surface per kelvin). Unlike $k$ it is not a pure material constant — it depends on flow speed and geometry, because faster flow thins the insulating still-fluid boundary layer. Still air $h \approx 8$ , fan air $h \approx 60$ , water $h \approx 1000$ .

The two building-block resistances (both derived in the parent note) use exactly these symbols:

Conduction (heat crawling through a solid): $R_{t h, cond} = \frac{d}{k A}$ — thickness $d$ over conductivity $k$ times area $A$ .
Convection (heat swept away by moving air/liquid): $R_{t h, conv} = \frac{1}{h A}$ — one over the heat-transfer coefficient $h$ times area $A$ .
Resistances in series add: $R_{t h, total} = R_{1} + R_{2} + \dots$ (because heat must cross each stage one after another, so their "pushes" $Δ T_{i}$ stack up).

Before the exercises, one picture to see why those two formulas take the shapes they do — the visual-first foundation the rest of the page stands on:

Figure — Heat dissipation and cooling solutions

Left (conduction): heat is squeezed through a solid slab. A thicker slab (bigger $d$ ) is a longer tunnel, so resistance grows with $d$ (top of the fraction). A wider slab (bigger $A$ ) offers more parallel lanes, so resistance shrinks with $A$ ; better material (bigger $k$ ) also shrinks it — hence $R = d / (k A)$ . Right (convection): heat leaves a surface into moving air across a thin still-air boundary layer. More surface $A$ or faster flow (bigger $h$ ) both help, and nothing "thick" is in the way, so $R = 1/ (h A)$ — no $d$ appears.

Now to the chain that the exercises actually solve, over and over:

Look at the chain: the red CHIP on the left is the heat source (power $P$ , the "current"). The three coloured boxes are the series resistances $R_{1}, R_{2}, R_{3}$ from Exercise 2.1 — junction-to-case, paste, heatsink-to-air. The blue AIR box on the right is $T_{amb}$ , the floor everything sits above. The grey arrows show heat marching left-to-right through every box in turn: that "in turn" is exactly why the resistances add. The temperature climbs by $Δ T_{i} = P R_{i}$ across each box, and the labels at the bottom show those drops summing to the 90 °C rise and 115 °C junction temperature you will compute in Exercise 2.1.

Level 1 — Recognition

Exercise 1.1

A cooling path has $R_{t h} = 0.6$ °C/W and carries $P = 150$ W. What is the temperature rise $Δ T$ across it?

Recall Solution

Use the master law directly — nothing to rearrange. $Δ T = P \times R_{t h} = 150 \times 0.6 = 90 °C$ What we did: multiplied flow by resistance. Why: this is the thermal version of Ohm's law $V = I R$ ; $R_{t h}$ tells you how many °C each watt buys you. Answer: 90 °C.

Exercise 1.2

Copper has thermal conductivity $k = 400$ W/m·K, aluminium $k = 205$ W/m·K. For the same slab (same $d$ , same $A$ ), which has the lower conduction resistance, and by roughly what factor?

Recall Solution

$R_{t h, cond} = \frac{d}{k A}$ . With $d$ and $A$ fixed, resistance is inversely proportional to $k$ . Bigger $k$ ⇒ smaller $R_{t h}$ . $\frac{R _{Al}}{R _{Cu}} = \frac{k _{Cu}}{k _{Al}} = \frac{400}{205} \approx 1.95$ Copper wins, with about $1.95 \times$ (≈ 2×) lower resistance. That's exactly why copper appears in high-end heatsink bases. See Heat pipes for the extreme version of this idea. Answer: copper, ≈ 2× lower.

Exercise 1.3

Match each mechanism to its dominant role in a desktop CPU cooler: (a) chip die → heatsink base, (b) fins → surrounding air, (c) fins → distant wall.

Recall Solution

(a) chip die → heatsink base: conduction (heat travels through solid metal).
(b) fins → air: convection (moving air carries heat off the surfaces).
(c) fins → distant wall: radiation (electromagnetic — under 5 % of the total below ~100 °C, so negligible here; governed by $Q = ε σ A (T_{s}^{4} - T_{amb}^{4})$ , see the footnote above). Answer: a=conduction, b=convection, c=radiation.

Level 2 — Application

Exercise 2.1

A CPU dissipates 150 W. The path is junction-to-case $R_{1} = 0.2$ °C/W, thermal paste $R_{2} = 0.1$ °C/W, heatsink-to-air $R_{3} = 0.3$ °C/W. Ambient air is 25 °C. Find the junction temperature $T_{j}$ .

Recall Solution

Step 1 — add series resistances. Heat must cross the chip package, then the paste, then the heatsink one after another — nothing skips a stage. When resistors sit end-to-end like that, each contributes its own temperature drop and they stack, so total resistance is the plain sum: $R_{t h, total} = 0.2 + 0.1 + 0.3 = 0.6 °C/W$ Step 2 — temperature rise. Apply the master law $Δ T = P R_{t h}$ , because that is precisely how many °C the total resistance costs at this power: $Δ T = P \times R_{t h, total} = 150 \times 0.6 = 90 °C$ Step 3 — add ambient. The 90 °C is a rise above the surrounding air, so to get the actual junction temperature $T_{j}$ we add the baseline $T_{amb}$ the whole stack sits on: $T_{j} = T_{amb} + Δ T = 25 + 90 = 115 °C$ Answer: 115 °C. Compare this against Thermal Design Power (TDP) limits — if $T_{j}$ exceeds the spec, Thermal throttling kicks in.

Exercise 2.2

A heatsink has effective surface area $A = 0.05$ m². Under natural convection $h = 8$ W/m²·K; with a fan $h = 60$ W/m²·K. Compute both convective resistances and the improvement factor.

Recall Solution

Use $R_{t h, conv} = \frac{1}{h A}$ , because the fin surfaces shed heat to air by convection and this is the resistance of exactly that stage. Natural: $R = \frac{1}{8 \times 0.05} = \frac{1}{0.4} = 2.5 °C/W$ Forced: $R = \frac{1}{60 \times 0.05} = \frac{1}{3.0} = 0.333 °C/W$ Improvement: $\frac{2.5}{0.333} = 7.5 \times$ Why so large? $h$ jumped 7.5× because the fan thins the insulating boundary layer of still air. Since $R \propto 1/ h$ , resistance drops by the same factor. Answer: 2.5 → 0.33 °C/W, a 7.5× improvement.

Exercise 2.3

You need to dissipate 80 W with the heatsink base no hotter than 90 °C, in 30 °C ambient. The chip-to-base path is already 0.25 °C/W. What heatsink-to-air resistance is required?

Recall Solution

Step 1 — temperature budget. Heat only flows on a difference, so the usable budget is the base cap minus ambient: the whole path may only rise $90 - 30 = 60$ °C. Step 2 — total resistance allowed. Rearrange the master law to $R = Δ T / P$ , because we now know the allowed rise and the power and want the resistance that fits: $R_{t h, total} = \frac{Δ T}{P} = \frac{60}{80} = 0.75 °C/W$ Step 3 — subtract the upstream part. Series resistances add, so whatever the chip-to-base stage already spends leaves the rest for the heatsink: $R_{t h, hs} = 0.75 - 0.25 = 0.5 °C/W$ Answer: 0.5 °C/W.

Level 3 — Analysis

Exercise 3.1

Two heatsinks are candidates for a 100 W chip in 25 °C air:

A: aluminium, $R_{t h, hs} = 0.35$ °C/W, plus paste 0.10, plus junction-to-case 0.15.
B: copper, $R_{t h, hs} = 0.28$ °C/W, plus paste 0.10, plus junction-to-case 0.15.

The chip must stay below $T_{j} = 95$ °C. Which cooler(s) pass, and what is the margin?

Recall Solution

Total resistances (sum the series stages, since heat crosses each in turn): $R_{A} = 0.15 + 0.10 + 0.35 = 0.60 °C/W, R_{B} = 0.15 + 0.10 + 0.28 = 0.53 °C/W$ Junction temperatures ( $T_{j} = 25 + 100 R$ , i.e. ambient plus $P R$ ): $T_{j, A} = 25 + 100 (0.60) = 85 °C$ $T_{j, B} = 25 + 100 (0.53) = 78 °C$ Both are below 95 °C, so both pass. Margins: A has $95 - 85 = 10$ °C; B has $95 - 78 = 17$ °C. Answer: both pass; A margin 10 °C, B margin 17 °C. B's copper base buys 7 °C of headroom — useful for Overclocking.

Exercise 3.2

Your fan-cooled heatsink gives $R_{t h} = 0.33$ °C/W for a 150 W chip in 25 °C air. You want $T_{j} \leq 80$ °C. Does it pass? If not, how much lower must $R_{t h}$ become?

Recall Solution

Current rise: $Δ T = 150 \times 0.33 = 49.5$ °C, so $T_{j} = 25 + 49.5 = 74.5$ °C. Since $74.5 \leq 80$ , it passes with $80 - 74.5 = 5.5$ °C of margin. (If it had failed, the required resistance would be $R = (80 - 25) /150 = 0.367$ °C/W — but we already beat that.) Answer: passes, 5.5 °C margin.

Exercise 3.3

A cold plate touches water with $h_{w} = 1000$ W/m²·K over contact area $A_{w} = 0.004$ m². The radiator dumps heat to air; its combined conductance is given as $h_{a} A_{a} = 3.0$ W/K — this is simply the product of the radiator's air-side heat-transfer coefficient $h_{a}$ and its total fin area $A_{a}$ , bundled into one number because only the product matters in $R = 1/ (h A)$ (a large radiator gets its low resistance from big $A_{a}$ , not from big $h_{a}$ ). Ignoring the small conduction inside metal, find the two convective resistances and the total. Which dominates?

Recall Solution

Water side ( $R = 1/ (h_{w} A_{w})$ , the convection resistance where water grabs heat off the cold plate): $R_{w} = \frac{1}{h _{w} A _{w}} = \frac{1}{1000 \times 0.004} = \frac{1}{4} = 0.25 °C/W$ Air (radiator) side — here we are already handed the product $h_{a} A_{a} = 3.0$ W/K, so: $R_{a} = \frac{1}{h _{a} A _{a}} = \frac{1}{3.0} = 0.333 °C/W$ Total (series, because heat crosses the water stage then the radiator stage in turn): $R_{t h} = 0.25 + 0.333 = 0.583 °C/W$ The air side dominates (0.333 > 0.25). This is the key insight of liquid cooling: water is superb at grabbing heat, but you still have to give it back to the air at the radiator — that's the bottleneck. See Case airflow and positive/negative pressure for feeding that radiator. Answer: $R_{w} = 0.25$ , $R_{a} = 0.33$ , total $0.58$ °C/W; air side dominates.

Level 4 — Synthesis

Exercise 4.1

Design check. A 200 W CPU must sit at $T_{j} \leq 90$ °C in a 35 °C room. Fixed path: junction-to-case 0.10, TIM 0.05. You must pick a heatsink. Available heatsink area is 0.06 m². What heat-transfer coefficient $h$ must the airflow deliver?

Recall Solution

Step 1 — total budget: $Δ T = 90 - 35 = 55$ °C, so $R_{t h, total} = \frac{55}{200} = 0.275 °C/W$ Step 2 — subtract fixed upstream: $R_{t h, hs} = 0.275 - 0.10 - 0.05 = 0.125 °C/W$ Step 3 — solve convection for $h$ : $R_{t h, hs} = \frac{1}{h A} \Rightarrow h = \frac{1}{R _{t h, hs} A}$ $h = \frac{1}{0.125 \times 0.06} = \frac{1}{0.0075} = 133.3 W/m^{2} \cdotK$ Interpretation: $h \approx 133$ is well above forced-air's typical 60–100, so a single stock fan won't do it — you need strong forced air (dense fins + high-static-pressure fan) or you drop to liquid cooling. Answer: $h \approx 133$ W/m²·K.

Exercise 4.2

Water carries heat away as a flowing coolant. Its temperature rise obeys $P = \overset{m}{˙} c_{p} Δ T_{water}$ , where $\overset{m}{˙}$ is mass flow rate (kg/s) and $c_{p} = 4180$ J/kg·K. A pump moves 1.5 L/min of water (density 1000 kg/m³) through a loop absorbing 250 W. How much does the water heat up as it passes the cold plate?

Recall Solution

Step 1 — mass flow rate. Convert 1.5 L/min to SI: $1.5$ L/min $= 1.5 \times 1 0^{- 3}$ m³ per 60 s $= 2.5 \times 1 0^{- 5}$ m³/s. Multiply by density $1000$ kg/m³: $\overset{m}{˙} = 1000 \times 2.5 \times 1 0^{- 5} = 0.025 kg/s$ Step 2 — rearrange for $Δ T_{water}$ . From $P = \overset{m}{˙} c_{p} Δ T_{water}$ , divide both sides by $\overset{m}{˙} c_{p}$ : $Δ T_{water} = \frac{P}{m ˙ c _{p}} = \frac{250}{0.025 \times 4180} = \frac{250}{104.5} = 2.39 °C$ Why so tiny? Water's huge heat capacity ( $c_{p} = 4180$ J/kg·K) means a small temperature climb absorbs a lot of energy — that's the whole appeal of liquid transport. The coolant barely warms, so the entire loop stays near-uniform temperature and the radiator can shed heat over a big, distant surface. Answer: ≈ 2.4 °C rise.

Level 5 — Mastery

Exercise 5.1

Full-system reliability trade-off. A server CPU (TDP 165 W) runs at 25 °C ambient. Two designs:

Design X: total $R_{t h} = 0.42$ °C/W (quiet fan).
Design Y: total $R_{t h} = 0.30$ °C/W (louder fan).

A well-known engineering rule of thumb states component lifetime roughly halves for every 10 °C rise in operating temperature. Taking Design X's junction temperature as the reference, estimate the relative lifetime of Design Y.

Recall Solution

Step 1 — junction temperatures. $T_{j, X} = 25 + 165 (0.42) = 25 + 69.3 = 94.3 °C$ $T_{j, Y} = 25 + 165 (0.30) = 25 + 49.5 = 74.5 °C$ Step 2 — temperature drop of Y vs X: $Δ = 94.3 - 74.5 = 19.8 °C cooler$ Step 3 — apply the halving rule. Lifetime multiplies by $2$ for every 10 °C cooler: $Lifetime ratio = 2^{Δ/10} = 2^{19.8/10} = 2^{1.98} \approx 3.94$ Interpretation: Design Y, though noisier, lasts roughly 4× longer in expected service life. This is the quantitative link between cooling and Reliability and MTBF — cooler silicon fails far less often. The noise cost buys a big reliability win. Answer: Design Y ≈ 3.9× the lifetime of Design X.

Exercise 5.2

Capstone. You inherit a machine that thermal-throttles under load. Measured: 180 W dissipation, ambient 28 °C, junction throttles at 100 °C. Current stack: junction-to-case 0.12, old dried paste 0.30, heatsink-to-air 0.35. (a) Confirm it throttles. (b) You replace paste (0.30 → 0.06) and add a better fan halving the heatsink resistance (0.35 → 0.175). Does throttling stop, and what's the new margin?

Recall Solution

(a) Current state. $R_{old} = 0.12 + 0.30 + 0.35 = 0.77 °C/W$ $T_{j} = 28 + 180 (0.77) = 28 + 138.6 = 166.6 °C$ That's far above the 100 °C throttle point — yes, it throttles hard (and is likely damaging the chip; see Thermal throttling). (b) After fixes. $R_{new} = 0.12 + 0.06 + 0.175 = 0.355 °C/W$ $T_{j} = 28 + 180 (0.355) = 28 + 63.9 = 91.9 °C$ Since $91.9 < 100$ , throttling stops, with a margin of $100 - 91.9 = 8.1$ °C. The lesson: the dried paste alone contributed $180 \times 0.30 = 54$ °C of rise — a single degraded interface can cook a whole system. TIM is not a detail. See PCB thermal vias and Power supply efficiency for the rest of the thermal picture. Answer: (a) throttles at 166.6 °C; (b) fixed, $T_{j} = 91.9$ °C, 8.1 °C margin.

Self-check clozes

The master law linking power, resistance and temperature rise

Δ T = P \times R_{t h}

Conduction resistance formula

R_{t h, cond} = d / (k A)

Convection resistance formula

R_{t h, conv} = 1/ (h A)

Series thermal resistances combine by

adding them

To get junction temperature from rise you must

add the ambient temperature

In liquid cooling the usual bottleneck stage is

the radiator (air side), not the water

Lifetime rule of thumb

roughly halves per 10 °C hotter

Why °C and K are interchangeable in these formulas

a 1 °C step equals a 1 K step (only differences appear)