# How much energy does it take to raise the temperature of an average …

• Someone can improve on my answer if they wish. This is my attempt to get an answer on the back of an envelope in less than 10 minutes. I ignore anything about walls leaking heat, or any objects in the room. You should be able to convince yourself that if you fill a room with desks (like a classroom) – they will take up considerably a small amount of volume compared to the air in the room itself (so the average temperature measured is probably biased towards the air temperature).

(dry air specific heat) cP=1.00J/g K

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${c}_{P}=1.00\text{J/g K}$

(dry air density)
ρ=1275g/m3

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$\rho =1275{\text{g/m}}^{3}$

(room area)
A=30m2

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$A=30{\text{m}}^{2}$

(room height)
h=3m

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$h=3\text{m}$

(mass of air)
m=ρAh

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$m=\rho Ah$

Change in temperature from 60F70F

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${60}^{\circ }\text{F}\to {70}^{\circ }\text{F}$

(I think this is from 15 Celsius to 22 Celsius?) so ΔT=7 K

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How much energy does it take to heat a room?

U=cPmΔT

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$U={c}_{P}m\mathrm{\Delta }T$

U=(1)(1275)(30)(3)(7)=800kJ

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$U=\left(1\right)\left(1275\right)\left(30\right)\left(3\right)\left(7\right)=800\text{kJ}$

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How long will it take with a 1000 Watt space heater?

P=UΔt

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$P=\frac{U}{\mathrm{\Delta }t}$

Δt=800 seconds15 minutes

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If this number was going to change because I made approximations, it will be because we require more energy than what I’ve listed here considering some leakage due to specific heat of walls, some conduction, and the fact that the room isn’t bare. I can’t imagine this factor being significant assuming the room is built pretty standard (throw in a fudge factor of like 5).

Compared to the other answers, I would say the heat capacity of the air in the room is just a smaller part of the equation. The heat capacity of the surfaces of its ceiling, floor and walls is much greater. Depending on the sort of materials that are used, this value can be anything in the range of 2 to 60 Wh/(m2*K). Say that the area of the surfaces combined is 115 m2, and that the average heat capacity is 5 Wh per m2 and degree K, and you want to heat this 10 degree F (5.5 K), then the surfaces would absorb about 3.2 kWh of heat. This is 3.2 hours of heating with the 1000 W heater.

This question is much more complex, as heat losses through conduction come into play. The temperature difference between the room and its surroundings (other rooms and the exterior), results in a increasing heat loss as the temperature rises. This would influence the time (and energy) needed to raise the room temperature. Another thing is ventilation, as the air inside can be changed several times during the transition to the higher temperature.

So, the best answer I can give, is it would take several hours (and kWh) to raise the temperature of this 300 sq foot room 10 degrees F.

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Mammals.

It was common in Europe in the Middle Ages to bring animals into the house. I used to be a castle guide on Heidelberg Castle, and that was one of the stories I used to tell the visitors.

There would be wooden insulation cladding on the walls, and straw on the floors, and the labourers and servants would sleep there, amidst goats and dogs and cats.

That construction was the reason for the fire that destroyed Heidelberg Castle, but I don’t think the goats were to blame.

Anyway, if you have a reasonably small, well-insulated …

Assuming you want to only increase the temperature of the ‘air’ in the room (and not the walls and other things), the calculation will use the specific heat capacity of air, the mass of air and the temperature difference to arrive at the amount of energy needed.

Assuming the height of the room is about 10 ft tall, the total volume of the room is 10 x 300 = 3000 cu.ft.

Amount of energy needed to raise temperature is given by the formula:

specific heat capacity of substance x mass of substance x temperature difference

1. Specific heat capacity of air is 1.006 kJ/kgC

2. mass of air:

density of air is 0.036 kg/cu.ft.

mass = density x volume = 0.037 x 3000 = 111KG

3. Temperature differential = 70C – 60C = 10C

energy needed = 1.006 x 111 x 10 = 1116 KJ

Power is energy transferred / second. That means, your 1000W space heater can transfer 1000J of energy / second. So time taken to heat up this space will be:

1116000 / 1000 = 1116 seconds. In other words: 18.6 minutes.

This of course assumes that the output power of your heater is 1000W. If the ‘input’ is 1000W, you may want to find out the efficiency of your heater and then arrive at the actual output power of the heater by the formula output power = input power x efficiency.

assumption : well insulated, no heat loss

1. First of all, i will calculate the energy needed for each part of the room to be raised from 60->70.

the parts consist of :

• Wall
• Floor
• Ceiling
• Air
• +Furniture etc

2. Find/estimate the volume of each part and find its mass by knowing its density

3. Find the specific heat (cp) of each part

4. Calculate the heat needed for each part ( Q = m. cp. dT) and sum it up

Now, you will have the total heat needed

5. Divide that heat by the heat supply, and voila

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Other answers provide calculations about specific heat (energy per Celsius) for present materials.

I will just point out to heat a well isolated volume of air – you need not so much energy.

To heat the walls and furnite, much more.

But to answer the question, to actually raise the temperature of the room, we must account to heat loss to outside during the process of reaching the increased temperature. In a badly isolated room and a slow heating mechanism, this is a lot.

which brings us to my main practical point about te relevance of these answers: if your room is perfectly isolated, then after bringing it up to the temperature you wish it will stay there and not cool over time. This is of course impossible without further heating, and in fact the only energy expended to heating on the long run goes towards heat leaking outside.

So to summarize, I want to point out that to keep your room 10 degrees warmer, the energy needed to heat all the air and objects in the room is irrelevant. It will make a minor difference, some energy during initial heating, but to then keep the room at that temperature means to heat enough to counter leakage, and this leakage will on the long run be virtually all of your energy demand. It won’t even matter what it’s made of or how big it is – except through how heat conduction of each material and surface area affect leakage to outside. That is of course before we start talking about ventilation 😀 which in turn is another huge factor in keeping the room at warmer temperature, despite the low specific heat of air: heat transfer by heat diffusion (conduction) is much slower than by movement of the heat-carrying material such as airflows (convection). In fact much leakage from a room is already facilitated by air currents: hot air is cooled by say a single glass layer window connected to the cold outside, the cooled air flows down and pushes air forwards, the warmed up furniture and heater warm up this air and it rises up and pushes warm air to the window etc. this leads to a lot higher heat loss by leakage through the window. Pumping cold air in and warm air out, such as for ventilation, will impose another serious energy expenditure to maintain the 10 degrees higher temperature. Which makes the specific heat calculations in other answers part of the 100% correct answer but also not-so-useful for any practical purposes.

i’m going on a gut feeling here…but i’m gonna say .87 cents…that’s with all previous assumptions but only getting it there and not sustaining it

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First, you determine the outdoor temperature, windspeed, barometric pressure, relative humidity, and solar angle. Next, you determine the indoor starting temperature. Then you painstakingly calculate the rate of heat loss from the building during the hours in which you plan to raise the temperature, taking solar gain and convection losses into account, and being sure to accurately determine the effective R-value of the insulation (if any) in the walls and cap and the total losses due to convection and infiltration / exfiltration past all doors and windows. Be sure to figure whether you expect any people to pass through any exterior doors during your test, and if so, how long the door will be open and the windspeed while it’s open. While you’re at it, don’t forget to consider whether the house is on a slab or piers or a crawspace or a basement… and whether floor insulation has anything to do with your calculations.

Looks simple, but it’s a lot more complex than the HVAC salesman would have you believe. Their calculations can be off by hundreds of percent.

Just experiment by picking a typical chilly night when the forecast is for stable overnight temperatures. Read the outdoor temperature, then read the indoor temperature at about 2: 00 AM. Start heating the house with two 1,000-watt electric milkhouse heaters, understanding that each of those heaters produces 3412 BTU/hr so two produce 6824 BTU/hr. Watch the indoor temperature rise (a) until it stabilizes or (b) until sunrise. That should tell you how quickly 3412 BTU/hr will elevate the temperature inside that house by 5 degrees under that specific set of circumstances… and give you everything you must know to calculate worst-case heating energy demand for that house.

No, if the space is enclosed, then a fan will not reduce the temperature. In fact, it will increase it slightly. To reduce temperature, the fan must be able to move air from a cooler place into the room in question, which it cannot do if the room is sealed.

However, if you are in a sealed room, a fan can still provide you with comfort by increasing the rate of evaporation from your skin, which cools your body.

Edit: If there is a conductive heat sink inside the room, such as a cold floor, then the improved mixing of the air will also reduce the average temperature of the room.

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How do you calculate the amount of energy that is needed to heat up 1m³ of sandstone from 10°C to 20°C?

Look up the density of sandstone (a search engine will find a suitable site)

You should be provided with a range, as sandstone is an aggregate of multiple constituents and the composition and aggregation will both vary. You may have been given a standard way to present this; if so use it. If not, I suggest that you calculate for both ends of the range and give your answer as minval – maxval.

Calculate the mass of 1 cubic metre of sandstone.

Look up the specific heat of sandstone (the room temperature value should be good enough for this)

Multiply this by the mass of the sandstone to get the heat capacity per degree

Multiply by the temperature change to get the required energy.

Make sure you use the same units throughout. The answer should be more than one megajoule.

The specific heat capacity of a substance tells you exactly this: how much energy is required to raise a certain mass of the substance by one degree. Accordingly, the units are kJ/kg.K, or “kilojoules per kilogram per Kelvin”.

The specific heat capacity of air at 20 deg. C and 1 atm pressure is 1.005 and 1 m^3 air has a mass of 1.205 kg, so you’ll need to multiply the specific heat capacity by this.

You should be able to work out the answer from here.

It takes more energy to heat the room, but the difference is ridiculously small for small intervals of temperature such as 1 degree.

Even though the energy consumption will be thermodynamically or ideally almost the same, heating something up requires a little more than cooling down. This is simply because the specific heat capacity of any material/substance (including the air in the room) always increases with temperature.

The specific heat capacity is an intrinsic property of a material that basically establishes how much heat it can take up before increasing temperature. It is related to the energy that must be given to a certain material in order to increase the inner kinetic and potential energies of the molecules. Proof of this dependence on temperature may be found in almost any thermodynamics/statistical mechanics book.

If you remember the basics of thermodynamics, we can write the energy that is given to a system by Q=mcΔT

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$Q=mc\mathrm{\Delta }T$

. If we want to write it more accurately, we do so by δQ=mc dT

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. The integration of this equation will say that the magnitude of Q is higher when dT > 0, because c is always increasing.