If you're worried about inductive EMF we can calculate that. Let's assume that we go from 10 A to zero in 1 ms. This seems a reasonable minimum time value as the contactor is mechanical and the mains is only at peak value for a fraction of each half-cycle anyway. Let's also assume that your value of 8 uH is correct. (I'm not going to check.)
$$ V = L \frac {di}{dt} = 8\times 10^{-6} \frac{10}{10^{-3}} = 8 \times 10 ^{-2} = 80~mV $$
I don't think you need worry about that.
Figure 1. (left) Pristine contacts from a relay
(right) The nearly destroyed contacts from a relay operated under power for nearly 100,000 cycles. Source: Wikipedia Arc Suppression.
The contact wear will come from the arcing and decreased contact area and pressure as the contacts open and close. Additional heat will be generated at this time.
Update (after OP supplied datasheet):
Figure 2. Extract from datasheet.
That relay is too small despite the manufacturer's specification. How do I know? It failed!
Note that with 100 mΩ contact resistance (and not 100 MΩ as stated) that at 10 A you will have \$ P = I^2R = 10^2 \times 0.1 = 10~W \$ dissipated by the contacts.
That contact resistance is quoted at 6 V so that's probably worst case but it gives some idea of the hardship the device endures
Secondly, since it's a DC coil I presume you are switching it electronically and have a snubber diode across the coil. This will slow down the release and increase the problem dramatically.
Tell your colleague that inductance is not the problem here.
OP's comment:
Anyway, I cannot agree with thansistor's [sic] prove, because calculations hasn't took into account the real value of inductance, which is unknown for the complete electrical circuit...
OK. What is the supply impedance? A quick web search brought me to an audio enthusiast site, Acoustica. (These guys worry about everything.)
One of the factors frequently overlooked by the hifi comics is that the electrical mains is not infinitely powerful; more specifically it can exhibit significant reluctance to supply enough electrons. Sometimes the impression is left that you could have the entire output of SIzewell B [a British nuclear power station] on tap if you buy a big enough power cord. In this year's colour, of course....
It ain't necessarily so; it ain't even close. IEC725:1981 models the European domestic mains supply as having an impedance of (0.4+j0.25)ohms. Surprisingly, measurement shows that the UK agrees with or somewhat betters the model on the whole, at something like (0.25+j0.23) ohms. What does that really mean? It's a definition which include the real resistance of the distribution network and wiring up to the socket, estimated as 0.25ohms, plus the reactive (j) component, allowing for stray capacitance and inductance of the wiring. The impedance is dominated by inductance, and equates to roughly 0.23ohms in the UK, for a net supply impedance of nearly 0.5ohms at 50Hz.
I haven't verified these numbers but let's take the inductance of the mains at your water heater supply as having a 0.25 Ω inductive impedance. We can calculate the inductance as follows:
$$ L = \frac {X_L}{2 \pi f} = \frac {0.25}{2 \pi 50} = 8 \times 10^{-4}~H $$
Plugging these back into our first equation and ignoring the \$ 8 \times 10^{-6} ~H \$ of the heater wiring we get:
$$ V = L \frac {di}{dt} = 8 \times 10^{-4} \frac{10}{10^{-3}} = 8~V $$
I think you can tell your colleague that we're still OK. Inductance isn't the problem.
I've never looked at this problem before so please let me know if I have missed something.
Links
There is another reference to IEC 725:1981 in The Influence of Source Impedance in Electrical Characterization of Solid State LOighting Sources by D. Zhao and G. Rietveld, VSL, Dutch Metrology Institute below equation (6) on page 3. This uses the same 0.4+0.25j Ω as the Acoustica source.
