Everything has a "specific heat capacity." They are derived expirementally (I believe) and you just have to look them up.
Heat capacity of H2O...
Specific heat capacity of water is ~4.18J/g·°K.
(It takes a little more than 4 joules of energy to increase the temp of 1g of water by 1° Celsius)
A mix of gases typically referred to as "air," has a specific heat capacity of ~1J/g·°K.
(It takes about 1 joule of energy to raise the temp of 1g of air by 1° Celsius)
It takes 4× as much energy to raise 1g of water by 1°C as it does 1g of air. If you add 100J of energy to 1g of water, and 100J of energy to 1g of "air," the air will have ~4× the increase in temp compared to the water, if they both started at the same temp.
Ex. 1
10L of water in a pot. Stove is on for 1.5min and is 500W, and all the energy is transferred to the water. What is the final water temp?
Given:
1L = 1,000mL
Heat capacity of H2O = 4.18J/g·°K
1mL of H2O = 1g of H2O
(W)att = (J)oule/(s)ec
Find grams of H2O...
10L × 1,000mL = 10,000mL; 10,000g of H2O
Find total joules...
500W = 500J/s
500J/s × 60sec/min × 1.5min = 45,000J total
Find how many joules per gram were added...
45,000J ÷ 10,000g of H2O = 4.5J/g
Find the temp increase...
4.5J/g ÷ 4.18J/g·°K =
1.07°C increase
Ex. 2
The pot of water and stove consume 1.0m3 of space and is in a square cube of a kitchen with dimensions of 3m×3m×2.5m. Assuming an air density of 1.225kg/m3 at STP, and assuming the walls and stove ect don't absorb any energy, what effect did the increase in water temp have on the air?
Given:
Air density = 1.225kg/m3
Specific Heat = 1.01J/g·°K
Stove & pot = 1m3
Find air volume...
3m × 3m × 2.5m = 22.5m3
22.5m3 - 1m3 = 21.5m3 of air
Find air mass...
21.5m3 × 1.225kg/m3 = 26.335kg of air
Together the 2 items make up the entire energy absorbing mass of the room, air, and the pot of water. 10kg of H2O and 26.3kg of air, or 36.3kg of total mass. The total amount of energy is dispersed about both masses in this example, as opposed to only the water in the first example. All 45kJ go into all 36.3kg, but the energy won't be split evenly between the 2 masses, just that the 2 masses together will absorb all 45kJ.
10,000g H2O × 4.18J/g·°K = 41,800J/°C
(It takes 41.8kJ to raise 10kg of H2O 1°C)
26,337g air × 1.01J/g·°K = 26,600J/°C
(It takes 26.6kJ to raise 26.3kg of air 1°C)
Because both masses are acting and absorbing the energy being input, we have to add the total J needed to raise each mass by 1°, together.
41,800J/°C + 26,600J/°C = 68,400J/°C
If we added 26.6kJ to 26.3kg of air, the 26.3kg of air would raise by 1°C, and if we added 41.8kJ to 10kg of H2O, the 10kg of H2O would raise by 1°C. If we combined the two substances together we don't need to add the individual temp raises, because things of the same temp (each raised by 1°C) don't change each other's temp. Each mass gained 1°C, and together in total the entire mass then also had to gain only, and exactly, 1°C. This means that it takes a total of 68.4kJ to raise 26.3kg of air + 10kg of H2O, by 1°C. Or, said another way, it takes 68.4kJ to raise a 36.3kg mixture of air & water by 1°C. Each mass independently requires its specific amount of energy to raise itself 1°C, but we have to add all the required energy amounts needed to raise each mass by 1°C respectively, of all the masses present, to determine the total amount of energy needed to raise the entire system (entire mass) by 1°C. And because over time the temps will equalize to a steady state temp, if we find total system temp change, we have then also found the individual mass temp changes of each substance too.
Solution...
45,000J ÷ 68,400J/°C =
0.657°C increase in total system temp from adding 500W of energy for 1.5min in a room with an air volume of 21.5m3 & 10L of H2O.
So the water gained 1.07°C, then after it was put into the kitchen the kitchen air absorbed some heat from the water, which resulted in the water reducing its temp, so that finally each mass had a net temp increase of 0.657°C.
A 0.657°C increase of 10kg H2O happens from absorbing 27,462J (41,800J/°C × 0.657°C), and a 0.657°C increase of 26.3kg of air happens from absorbing 17,476J (26,600J/°C × 0.657°C), and together the total energy absorbed to raise each by 0.657°C is 45,000J. The energy is dispersed unevenly amongst the 2 masses, but to an end result where each substance has the same temp increase.
This is a basic explanation but in real world you work more with rates or fluxes, so you calculate power going into the various masses as well as the power exiting out of the various masses and the excess energy present once the system is at equilibrium (or at any point in time desired), ie system temp increase. You have to account for thermal conductivities, and air flow patterns ect to be exact and/or to calculate instantaneous temperatures. Though you can still get a basic idea with the above examples, and you can create a spreadaheet to plot out your system a little better. Water is fairly dense and has pretty high heat capacity, so if you use a 55gal drum, you'll most likely have much more water than air (in terms of mass, ie g or kg, not volume), and a significant portion of the rooms mass will then be water, which means your estimates based soley on the water & air volume could semi-closely reflect the entire system. The difference in calculated temp vs observed temp is going to be due to the energy escaping out (or entering) the room, or excess mass unaccounted for (ie walls or ceiling/floor absorbing heat, conducting, and re-emitting on the other side, buckets, plants, ect).
