# Why does my house seem to warm faster in summer than it cools in winter? [closed]

In summers when we switch off the air conditioner, the room seems to instantly get hot again. But in winter, when we switch off the heater the room seems to remain hot for some time. Why this difference?

## closed as unclear what you're asking by ACuriousMind♦, HDE 226868, John Rennie, Kyle Kanos, Sebastian RieseOct 27 '15 at 15:47

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• This hasn't been my experience. In fact yesterday I came home from work and turned off my air-conditioner and hours later my girlfriend came over and said it was too cold! It depends on many factors: air flow, sun light intensity, location and area of windows, temperature difference inside versus outside... – DavePhD Jun 13 '14 at 12:48
• Related search: physics.stackexchange.com/search?q=is%3Aq+temperature+ac – Qmechanic Jul 17 '14 at 10:05
• Without a lot more information about the situation, this is a guessing game, not a well-defined physics question. – ACuriousMind Oct 26 '15 at 22:35
• Voting to re-open for no reason other than that I was about to ask exactly the same question myself. In my experience, rooms feel hot only a few minutes after turning off the air con in summer, while they may still feel warm hours after turning off the heating in winter. This might be to do with humidity, or it might be to do with the $T^4$ law (though it happens at night just as much as during the day). It's clearly a physics question, and the answer is not obvious. – Nathaniel May 29 '16 at 11:16
• (Relevant information, at least in my case: I live in Japan, where summers are very humid, winters are very dry, and insulation often leaves a lot to be desired.) – Nathaniel May 29 '16 at 11:18

I think part of the perceptual difference has to do with humidity. The human body doesn't really feel the temperature we read off of our thermometers. The thermodynamics of the human body are complicated, but people have designed various scales that are supposed to measure "apparent temperature". One of these is the Canadian humidex, which is a complicated formula in general, but when linearized they say:

The humidity adjustment effectively amounts to one Fahrenheit degree for every millibar by which the partial pressure of water in the atmosphere exceeds 10 millibars.

Instead of working with partial pressure, let's consider relative humidity, which is just defined as the fraction of partial pressure with respect to saturation pressure: $$r \equiv \frac{P}{P_s}$$ so we could say (with $A$ for apparent temperature): $$dA \sim dT + \frac{ P_s }{ 1 \text{ mbar} } dr$$

So, let's try to consider how we expect the temperature or humidity to invade after we turn off the heater/cooler. Well, we know that the process for both should look like Newton's law of cooling as they are both invasion processes, so we should have something like $$dT = \frac{ \Delta T }{ \tau_Q} dt \qquad dr = \frac{ \Delta r }{ \tau_v } dt$$ where here, $t_Q$ is some characteristic time for heat invasion in a home, and $\tau_v$ is a characteristic time for water vapor invasion. using these in our approximation for how the apparent temperature depends on temperature an humidity, we get an equation for the rate change for the apparent temperature and compare this to a combined effective rate

$$\frac{dA}{dt} = \frac{\Delta T}{\tau_Q} \left( 1 + \left(\frac{P_s}{1 \text{ mbar}}\right) \frac{\Delta r}{\Delta T} \frac{\tau_Q}{\tau_v} \right) \equiv \frac{\Delta T}{\tau}$$ we see $$1 + \left(\frac{P_s}{1 \text{ mbar}}\right) \frac{\Delta r}{\Delta T} \frac{\tau_Q}{\tau_v} = \frac{\tau_Q}{\tau}$$

Now, to estimate. The difference is going to come from the fact that in the summer, especially in humid climes (which I am assuming you are from, given your observation), in the summer, their might be a 20 degree difference in temperature from inside to out, but inside is usually at about 30% humidity while outside is usually near 100%, so we'll take $\Delta r = 0.7$. I'm imagining Florida, so we'll take $\Delta T \sim 20\, {}^{\circ}F$. The vapor pressure of air at 21 celsius (70 F) is $P_s \sim 26 \text{ mbar}$. It remains to estimate the ratio of the time constant for vapor invasion versus heat. In this blog post, the author does a fairly detailed calculation of the heat flow in a model house, considering conduction, convection and radiation through walls, floor, ceiling, windows, doors, etc, as well as the heat flow do to air invasion, and discovers that they are roughly equal in contribution. Now, as for humidity invasion, we expect it is negligible through the walls and floor and the like, but should be present in any air invasion. This suggests that $\tau_Q / \tau_v \sim \frac 12$ which gives us for an effective time constant for the apparent temperature in the summer, considering the contribution of humidity we get

$$\tau \sim \frac{ \tau_Q }{ 1.5 }$$

Now, in winter, there is usually a very small difference in humidity from inside to outside, but still a $\sim 20\, {}^{\circ}F$ change (in Florida) so we expect $\tau \sim \tau_Q$, which suggests that your house should feel like its getting hotter some 50% faster in summer than it feels like its getting colder in winter, due to the effect humidity has on apparent temperature.

A window air conditioner is a small piece of equipment with relatively small heat capacity. But with heaters it depends, if it is electric it barely has much capacity and the room will feel quickly very cold once it is turned off. If your heater is steam radiator or forced water then it will take a long time for those to cool down. If the room has central forced air to heat or cool then probably it does not matter either way.

When your house heats up, it is receiving the contribution from the hot air outside, plus the sun's radiated heat. When cooling in the Winter, the factors are the cooler air and the radiation loss, which is not comparable to the sun's.

This question can be answered with a nice visual comparison.

Let's take a pan of water, and put it on the stove at full power. You periodically throw in an ice cube to keep it at room temperature. Obviously, the moment you stop doing this, your pan is going to heat up very quickly. Now, take that same pan and put it in a slightly colder place. Intuitively, you can see that it won't cool down as quickly as it heated up with a big stove.

The same goes for your house in summer. The sun is a massive radiator; your house is hit with about 1kW per square meter on a sunny day. It's only because a large portion is reflected that your airco unit is even slightly capable of cooling your house. Check your airco; let's say it has a cooling power of approx 3kW: it needs to take away 3kW of energy continuously, or else your room will heat up as if there were a heater of 3kW.

Now, how cold does it have to be outside to to need a 3kW heater? Typical R-values are around 6 °K m²/W, which need you need 6 degrees Kelvin (or Celcius) difference between the inside and the outside to lose one watt of power on a square meter of wall. Let's say it's -20°C outside (that's -4 Fahrenheit), and 20°C inside; i.e., 40°C difference. You'll need a whopping 450m² of wall area to need a 3kW heater. Admittedly, this neglects quite some things left and right, but it gets the point across.

TL;DR: your house is not a glowing hot fireball like the Sun, and will not lose heat quite as fast to its surroundings, as it gains heat by solar radiation.