Solution
Physics 213
Problem 3
Soldering
Week 3
One of the real-world thermodynamic issues that arises when assembling electronic circuits is
the need to prevent heat damage to the components. Specifically, one needs to take care not to
leave the soldering* iron too long on the connection, lest the wire and the attached component
heat up too much. Here we make some simple estimates involving this problem.
Assume
• the thermal conductivity of copper wire is (κ = 400 W/m-K)
• specific heat of copper is c = 386 J/kg-K
• density of copper is ρ = 8900 kg/m3
• the length of the wire is L = 5 mm
• the diameter of the wire is d = 1 mm
• the initial temperature is 20°C
• the soldering iron temperature is Tiron = 250°C (assumed constant)
• the component will “die” when the copper touching it reaches Tdeath = 200°C
1) What’s the mass of the copper wire?
m = ρvol = ρ*L*π(d/2)2 = 3.495 x 10-5 kg
2) What’s the heat capacity of the copper wire?
C = cm = 0.0135 J/K
3) How many joules must be deposited to raise the temperature of the entire wire from
20°C to 200°C?
Q=C δT = 0.0135 J/K (200-20) = 2.428 J
4) What is the thermal resistance Rth of the wire, given that the cross-sectional area A = π(d/2)2
= 7.85 x 10-7 m2?
Rth = (L/κA) =15.92 K/W
5) What is the initial heat flow (i.e., before the wire temperature has changed much)?
H = ΔT/R = (250-20)/15.92 = 14.45 W
NOTE: This is a different ΔT than in Part 3! Here it’s the full temperature range.
6) As a rough estimate for how much time τ you have – how long it takes for an unsafe amount
of thermal energy to diffuse through the wire – simply assume the heat flow stays constant, and
calculate how long it would take to deposit the energy needed to heat the wire to 200°C (i.e.,
using your result from before):
*
Note: The “l” is silent in this oft-mispronounced word, whose first syllable rhymes with “hot”.
τ ≈ Q/H = 2.428 J/14.45 W = 0.168 s
7) Write your previous result for the thermal diffusion time τ in terms of the (thermal) resistance
Rth and the (heat) capacity C of the wire. (Hopefully this ‘time constant’ looks familiar from
electrical RC circuits!) Then show that τ does not depend at all on the cross-sectional area A of
the wire, but only on the length L (and on ρ, c and κ, and the temperature differences).
This sort of quadratic dependence (τ ∝ L2) is typical of any random diffusion process. In this
problem it is mostly the hot electrons that are diffusing along the wire.
8) If we want to double τ, thereby doubling the time we have to solder before overheating the
component (at the other end of the wire from the soldering iron), what should we do to the length
of the wire?
Since τ ∝ L2, we can double the heating time by increasing the length by a factor of √2.
9) Finally, in order to estimate the time, we made the simplifying assumption that the heat flow
is constant as the wire heats up. What actually happens to the heat flow as the wire heats up, and
what is the effect (qualitatively) of this on the thermal diffusion time τ?
The rate of heat flow is proportional to the temperature gradient (i.e., ΔT). Therefore, as the wire
temperature approaches that of the soldering iron (ΔT decreases), the rate of heat flow decreases.
Consequently, it takes longer to deposit the thermal energy needed to increase the temperature of
the wire (a more detailed calculation would show that for the numbers given, the real τ would be
~0.33 s, about 1.5 x longer than our simple estimate).
FYI: How would we solve this problem without the simplifying assumption. Since the final
temperature of the wire actually changes a lot (in contrast to the HW problem on hypothermia,
where the temperature did not change much), we should really take this into account. As shown
in the lecture notes, the heat current into the wire is given by the differential equation:
,
with a solution
Twire(t) = Tiron + [Twire(t=0) – Tiron]e-t/τ, where τ = RthC.
Here is a sketch of this solution, starting from the initial wire temperature (20°C), and
asymptoting to the soldering iron temperature (250°C):
We want to know at what time will Twire(t) = Tdeath (250°C), i.e., how much time do we have
before the temperature of the wire (and therefore the attached sensitive component) reaches it’s
failure temperature. We can solve the solution Twire(t) = Tiron + [Twire(t=0) – Tiron]e-t/τ for t in
terms of τ, and plug in the various temperatures given in the problem:
t = -τ ln[(Tdeath – Tiron)/( Twire(t=0) – Tiron)] = -τ ln[(200-250)/(20-250)] = 1.526 τ
Plugging in our earlier value for τ = RthC , we find:
τ = Rth C = 0.215 s  t = 1.526 x 0.215 s = 0.328 s
Note that this quite a bit longer than the initial estimate in question 6), because the rate of heat
flow slows down as the wire temperature approaches that of the soldering iron.