How cold would a bowling ball near absolute zero make a room?

Question

If a bowling ball at absolute zero suddenly appeared in my room, how cold would the room get? Would I die? My initial estimation is that it one bowling ball at 0K (i.e., 300K below room temperature) would have about the same effect as 10 bowling balls at 270K (i.e., just below freezing, 30K below room temperature).

Answer 1

There are two parts to your question, and I'll address them individually, assuming that the room+ball system is an isolated system, i.e. no work is being done on the system, and no energy is being added / removed.

Part I: How cold will it get?

This depends on the size of the room, the materials that everything in the room is made of, the size and material of the ball, etc. Suppose, for the sake of this discussion, that there is only air in the 5m x 5m x 3m room, at 300K, and one "bowling ball" made of iron, having a mass of 7kg.

Iron has a specific heat of $C = 0.450 \mathrm{~kJ/kgK}$. Considering the ball heats up from $0 \mathrm{~K}$ to, say, $T \mathrm{~K}$. The amount of heat required for it to heat up will be: $Q = m_b C \Delta T = m_b C (T - 0)$

Using the density of air at 300K, we get the mass of air at this temperature to be $1.177 \times 75 = 88.275 \mathrm{~kg}$. The specific heat of the air at constant volume, $C_v \approx 0.716 \mathrm{~kJ/kgK}$. When the entire process has finished, the air in the room must be at the same temperature as the ball, $T \mathrm{~K}$. The heat removed from the air would be, $Q = m_{air} C_v \Delta T = m_{air} C_v (300 - T)$

Since the system is isolated, the heat removed from the air must be equal to the heat entering the ball. \begin{align} m_{air} C_v (300 - T) &= m_b C T \\ \therefore T(m_b C + m_{air} C_v) &= 300 m_{air} C_v \\ \therefore T &= \frac{300 m_{air} C_v}{m_b C + m_{air} C_v}\\ \therefore T &= \frac{300 \cdot 88.275 \cdot 0.716}{7 \cdot 0.45 + 88.275 \cdot 0.716} \\ \therefore T &= 285.758 \mathrm{~K} \end{align}

285.758 K is 12.6 C, which doesn't seem too uncomfortable.

Part II: Is one bowling ball @ 0K equivalent to 10 at 270K?

The heat equation for the balls is now $Q = 10 m_b C (T - 270)$

Plugging this into the equations: \begin{align} m_{air} C_v (300 - T) &= 10 m_b C (T - 270) \\ \therefore 300 m_{air} C_v - m_{air} C_v T &= 10 m_b C T - 2700 m_b C\\ \therefore T(10m_b C + m_{air} C_v) &= 300 m_{air} C_v + 2700 m_b C \\ \therefore T &= \frac{300 m_{air} C_v + 2700 m_b C}{10m_b C + m_{air} C_v}\\ \therefore T &= \frac{300 \cdot 88.275 \cdot 0.716 + 2700 \cdot 7 \cdot 0.450}{10 \cdot 7 \cdot 0.45 + 88.275 \cdot 0.716} \\ \therefore T &= 290.022 \mathrm{~K} \end{align}

Conclusion: The two cases aren't equivalent.

Note: This answer only takes into account the two end cases, i.e. the initial moment when the ball(s) are introduced into the room, and the final equilibrium state. Transient effects are ignored.

Note 2: This answer assumes the specific heat capacity of both air and iron remains constant over the entire temperature range, which, as Floris pointed out, is not the case. To consider a material having varying specific heat, the equations for $Q$ change to $Q = n_bm_b \int_{T_{min}}^T{C dT}$ and $Q = m_{air} \int_T^{T_{room}}{C_v dT}$, where $n_b$ is the number of balls, $T_{min}$ is the minimum temperature of the ball(s) and $T_{room}$ is the temperature of the air in the room before the ball(s) are introduced. The assumption that the $C_v$ of air is constant is justified because the temperature of the air in the room does not vary so much as to induce a great change in the $C_v$. This assumption would be invalid if the amount of air in the room were to reduce such that the difference in its initial and final temperatures turns out to be substantial.

Note 3: When you consider that the specific heat capacity of iron changes with temperature, there's a significant difference. Since I don't have an equation for the $C$ vs $T$ of iron, I'm using numeric data from here (hat-tip to Floris). The plot of $C$ vs $T$ looks like this:

Since I have numeric data, I took T = 286 K as a first approximation wrote an iterative Matlab script to calculate the residual, $\Delta = m_b \int_0^T C dT - m_{air} C_v (300 - T)$. At $T = 291.4455 \mathrm{~K}$, the residual was $\Delta = -4.2406 \times 10 ^ {-6}$

Using the same code for the second case, where the residual is $\Delta = 10m_b \int_{270}^T C dT - m_{air} C_v (300 - T)$. I get $T = 290.1689 \mathrm{~K}$ with a residual of $\Delta = -4.9199\times 10 ^ {-6}$

The conclusion is still the same, but the end temperatures are slightly different from the first calculation.