Extremely Quick Question (2 or 1 second)

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Heating tempered glass with small air bubbles, approximately a third of a millimeter in radius, is unlikely to cause the bubbles to explode, as they were formed at higher temperatures without issues. The primary concern is the potential for glass breakage due to thermal stress if heated too quickly. Trust in the manufacturer’s quality can provide additional assurance regarding the glass's integrity. Proper safety precautions, such as wearing goggles, are recommended when handling heated glass. Overall, the glass should withstand the intended heating without significant risk of explosion from the bubbles.
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Just a quick question about glass.. I bought a glass piece that is supposed to heat up to around 200 degrees celsius (it's tempered). I noticed that the glass has some bubbles in it. My question is..will the heat cause the the air in those bubbles to expand enough that they explode? Thx
 
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Well I guess it would depend ont he size of the air bubbles, the heat would probably cause it to expand, but wether it would cause the glass to crack I think it down to the size of the bubbles.
 
They have a radius of about a third of a millimeter..
 
The glass was formed at temperatures higher than 200C with the air bubbles already in there. So if they didn't explode when it was being cooled, they won't explode when it's heated up. What will cause it to break is stresses in the glass if you heat it up too fast, even if it is tempered. Take it easy.
 
If you can trust the company who produced the glass then you can fairly certainly trust the glass, if not...
Dick said:
Take it easy.

And wear goggles! :-p
 
thanks! :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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