Sometimes it happens that when you pour a boiling water into a glass jar, it cracks. Since glass is very hard material and resilient to pressure, the tension must be very high. Is it possible to estimate what forces / pressure / tension occur in such a moment?
Some supplementary questions:
The origin of the stress is differential thermal expansion. That is there is a temperature gradient imposed and warmer parts are trying to expand while the cool parts try to remain the same size.
From that it follows that adding hot liquid that has much more mass than the glass will impose larger stresses than adding a small quantity first.
To quantify this lets make some assumptions and do a basic calculation.
Assume the vessel has a uniform initial temperature of $T_i$, and that we pour in water at roughly boiling $T_f$. The walls are assumed to have thickness $t$, coefficient of thermal expansion $\alpha$, and shear modulus $G$.
We established the thermal gradient from the inside to the outside.
The most extreme case is when they innermost material is at the temperature of the liquid ($T_f$) and the outside remains at the ambient ($T_i$). Using an area element of initial linear dimension $l(T_i) = l_0 (1 + \alpha T_i)$, we get the strain on the system due to the thermal gradient of at least1 $$ \gamma \ge \frac{l(T_f) - l(T_i)}{t} = \frac{l_0 \alpha (T_f - T_i)}{t} \quad . $$ It may seem strange that the strain here depends on the size of the region, but the other way to look at this is that for fixed lateral size the strain varies inversely with thickness which is intuitively sensible.
Now, $G = \frac{\tau}{\gamma}$ where $\tau = \frac{F}{A}$ is the sheer stress, which implies that $$ F = A G \gamma = l(T_i) t G = l_0^2 G (1 + \alpha T_i) \alpha (T_f - T_i) \quad .$$ Note that the thickness has dropped out here, which justifies ignoring it's expansion. The sheer modulus of most glasses is between $20$ and $35 \text{ GPa}$ (we'll use 30), and the coefficients of linear expansion are near $8.5 \times 10^{-6} \text{ K}^{-1}$. Taking $T_i = 290\text{ K}$, and $T_f = 370\text{ K}$, we have enough to calculate the pressure: $$ P \approx 2 \times 10^7\text{ Pa} \quad .$$
1 The strain could be locally greater than this, if the temperature gradient is not uniform.
