<s>[INST] The density of water at $50^\circ\text{C}$ is $988 \text{ kg/m}^3$. The volume of a submerged stone equals the volume of the water displaced by the stone. The buoyancy force has magnitude $F_b = \rho V g$, where $\rho$ is the mass per unit volume of water and $V$ is the volume of water displaced. Since the stone is submerged, the weight of the stone is $F_g = m g = \rho_{stone} V g$, where $\rho_{stone}$ is the average density of the stone. The ratio of the buoyant force to the weight of the stone is $$\frac{F_b}{F_g} = \frac{\rho V g}{\rho_{stone} V g} = \frac{\rho}{\rho_{stone}}$$ We need $\rho_{stone} > \rho$. You throw both the stone and the water into outer space, far from any gravitational fields. Since the stone has a single value of volume, we know that expansion occurs as temperature increases. Thus the stone will expand from $50^\circ\text C$ to $90^\circ\text C$. Since $\rho = m/V$, we can write the change in volume of the stone as: \begin{align*}
\frac{\Delta V}{V} &= 3 \alpha \Delta T \\
&= 3 \times 4 \times 10^{-6} \times 40 = 4.80 \times 10^{-4}
\end{align*} where $\alpha$ is the average coefficient of thermal expansion of the stone.

We can choose $V = 3 \Delta V$. The expansion will be a change to$ 1.000480 \times V$. The $V$ used will be $V = 0.06796 \text{ m}^3$. Using this expansion, the value for $\rho_s$ such that $\rho/\rho_s = 0.54$ is $\rho_s = 1133.52 \text{ kg/m}^3$ 

The elastic modulus for the stone is difficult to calculate, however we can get the volume using the equation for bulk modulus:
$$K = -\frac{\Delta P}{\Delta V/V}$$ Filling out the calculations using $\Delta P = 1.01325\times10^5 \text{ Pa}$, $\rho = 1133.52 \text{ kg/m}^3$, and $\frac{\Delta V}{V} = 4.80\times 10^{-4}$ yields an answer that $\Delta P = K \frac{\Delta V}{V}$ results in $K = 2.11 \times 10^{10} \text{ Pa}$.

If $\frac{\Delta V}{V} = 6.375\times 10^{-5}$ results in a density that is greater than $988\text{ kg/m}^3$, then the stone is on the verge of being crushed, and $V$ is given by $V = \frac{m}{\rho}$. The equation becomes
$$K = \frac{\Delta P}{\Delta V/V} = - V \frac{\Delta \rho}{\Delta V} = -\rho \frac{\Delta (\ln\rho)}{\Delta (\ln V)} = \rho \frac{\Delta (\ln V)}{\Delta (\ln \rho)}$$ and substituting values, the answer results in
$$
K = -\frac{\ln 988 - \ln 1035.012}{\ln (1 + 6.375\times 10^{-5}) - \ln (1 + 4.80\times 10^{-4})}
$$
The results all show that for this instance $K>6.4\times10^9\text{ Pa}$. Then the question I have is how do all the previous calculations result in a value of K?[/INST]