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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).

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Next to nothing would happen that wouldn't happen for a bowling ball at the temperature of liquid nitrogen (and that would be little enough). Heat capacity near absolute zero is very small (approaching zero), which means that cooling down an object ever further doesn't do much, at all, in terms of the heat that it can absorb. At most, tho cold would freeze your skin, if you were to touch it... which would be foolish, painful and perfectly survivable. –  CuriousOne Sep 1 at 23:28
    
Er ... it (the ball) would warm up? I mean, your room would get chilly for a few minutes, but nothing catastrophic is going to happen unless you insist on cuddling up to it and that will hurt enough to convince you not to. –  dmckee Sep 1 at 23:31
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It depends mostly on the velocity of the bowling ball when it appears, and your position relative to it... :-) Seriously, thermal conductivity of bowling ball is low, so the effect would be small - smaller than the effect of 10 bowling balls that are only 30 K below room temperature. Not only would the latter have slightly more relative heat capacity, but more importantly they would be able to give off their heat 10x more efficiently (greater total area). –  Floris Sep 1 at 23:32
    
Simple thermodynamics. –  mcodesmart Sep 2 at 19:57

1 Answer 1

up vote 8 down vote accepted

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.

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You have an error: it should be $T(10m_bC+m_{air}C_v) =$... in the third line of part II. –  LDC3 Sep 2 at 3:26
    
My calculator comes up with 290.01 K. –  LDC3 Sep 2 at 3:35
    
Thanks @LDC3. I'm logged in on my phone, so it is slightly cumbersome to scroll through all that Latex. I've fixed the errors now. –  Pranav Hosangadi Sep 2 at 3:37
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Please. When you are estimating quantities, don't use so many (in)significant figures. Especially since your heat capacity is WAY OFF. See imgur.com/3aPjXGc for a graph of the values for iron from 1K to 298 K (taken from nist.gov/data/PDFfiles/jpcrd298.pdf). –  Floris Sep 2 at 6:55
    
@Floris, thanks for the tip. I wrote that answer last night from my phone, so it was extremely simplified. I've now added calculations considering the changing $C$. –  Pranav Hosangadi Sep 2 at 21:26

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