Let’s say you freeze (6) gallon milk jugs of water in your freezer. Before you go to work in the morning you put 3 on the floor behind the drivers seat and 3 on the floor behind the passenger seat. You make your 30 minute commute to work and then leave your car parked in direct sunlight with the windows up for 8 hours on a 90 degree day.
When you leave at 5 o’clock is your car any cooler inside versus not having the ice in there?
At some point it may be slightly cooler, but after 8 hours of direct sunlight in the car, the temp would have equalized and be just as hot as it would have been without the jugs. That’s my immediate guess anyways, I’ll do a thermo calc. later if time permits.
Would that mean the ice would have turned completely back into water? Could a hot car melt it that fast?
I agree with Dorjän.
6 gallons of ice would require about 8MJ of energy to melt, plus another 2MJ or so to bring up to room temperature. If your car has a cross-sectional area of 3 m[sup]2[/sup] (as seen from the sun), it would take less than an hour to receive that much energy.
After eight hours in direct heat… yes I would think the water would have melted.
Not all of which would be transferred to the milk jugs, but I agree, eight hours later, you got seriously hot water, and maybe a popped jug or two.
Tris
You would have more effect by NOT having the windows up. Leaving them open just a bit at the top will allow heated air to vent out of the car. (Note that this is not a cooling effect, it’s just reducing the heat buildup inside the car.) So it won’t make it cooler, but it will tend to keep it at the same temp as the outside air in the parking lot. If it’s all closed up, all the heat the sun pours into the car all day long will just build up in there, and the inside of the car will be significantly hotter than the outside.
You can enhance this effect by increasing the ventilation. There are small fans, designed to easily mount on car windows, and powered by solar panels attached to the roof. These do an even better job of keeping the car no hotter than the outside.
True. But I think it’s correct to say the ice would only cancel out 1 hour worth of sunlight. It’s equivalent to the car spending 1 hour of the day in the shade. (Or getting to work 1 hour late.)
Plus, you’ve got to worry about the puddle of condensation in the car. And the electricity required to make 6 gallons of ice is not insignificant either.
So how long would you figure it would take in the car to completely melt the 6 jugs and bring them to air temp?
Whatever the answer would be, if you took half that amount of time and checked the cars interior temp, any cooler than normal no ice car?
Another related question: Does a freezer use more electricity freezing things than just keeping them frozen?
I was doing all kinds of complicated math figuring out how much energy would be involved in heating various things up, but here’s a back-of-the-envelope kinda thought:
Specific heat of iron is 0.11 cal/gm-C. Specific heat of water is 1 cal/gm-C.
Let’s say your car is 4000 lbs (1800kg) and 6 gallons of ice is about 21kg of water.
So it takes about 200kilocal to raise the temp of the metal by 1 degree C, and 220kilocal to raise the temp of the (metal plus water) by 1 degree C. About a 10% increase, measureable but not astounding.
If the car begins at 15C (about 60F) and the water at 0C then it’ll take about 5400 kilocal to heat the whole system up to 38C (100F) - I’m sure that the inside will get hotter but that’s the number I happened to use. If you didn’t have the water it’d need about 4600 kilocals.
Now let’s look at how much energy is in sunlight - one site said an average of about 1000 watts/square meter on a cloudless day at high noon. One watt-hour is 860 cal.
Let’s assume that your car has an area (viewed from directly overhead) of 5’x15’ which is about 7 square meters.
That means your car is exposed to over 6000 kilocal every hour. It will take the car a little longer to get enough energy to heat up to 38C if you have the ice in there, but not much (needs about an extra 17% energy with the water in there).
So in the first hour your car is exposed to enough heat energy to bring the water up to temp along with the car. From that point on it just has to hold equilibrium temp. Maybe it’d be a little cooler over a short time period (less than an hour) but in the long run (8 hours), no difference.
Like I said, back of the envelope. I’ve ignored a whole mess of stuff (% energy actually absorbed by the car, heat to turn ice into water at 0C, etc) but I think this gets us in the ballpark.
Yes, obviously.
A non-frozen thing is at some temperature above 32ºF, probably room temperature. To freeze it, you need to remove heat until it is down to 32ºF or below. (For most freezers, they are typically set at about 0ºF.)
Once it is frozen, all that is needed is to keep it frozen (below 32ºF). That just means removing the heat that leaks into it, from outside the freezer. What it takes to do that depends on the quality of the insulation on the freezer, how often the door is opened, and how often something non-frozen is added to the freezer. But the energy needed to do this is quite a bit less than that needed to freeze it in the first place.
A solid block without forced air will have poor heat transfer. Even if you had replacement frozen bottles that you would substitute as each the blocks melted, I suspect the opening and closing of the door would reduce the temperature substantially more than the block.
Actually they make a solar powered fan in a plastic strip that is designed fit at the top of a car window and be locked in place by rolling up the glass, it has slides that adjust to the opening. The solar panels are on the outside of the strip. One of these makes a noticable difference, if you used two, or maybe four, and a sunscreen in the windshield, I think it would make an even bigger difference.