Two objects in contact reach thermal equilibrium when they have the same temperature and there is no net flow of heat between them. Heat flows from the hotter object to the cooler one until that happens.
This lesson is part of heat. It prepares you for using specific heat capacity.
What should the explanation say?
A full answer has three statements.
- Heat flows from the hotter to the colder object.
- The hot object cools and the cold one warms up.
- At equal temperature there is no net heat flow, so they are in thermal equilibrium.
Leaving out statement 3 is the usual reason marks are lost.
Worked example: hot metal in water
A 0.50 kg metal block at 100°C is placed in 0.20 kg of water at 20°C. The specific heat capacity of the metal is 400 J kg⁻¹ °C⁻¹, and for water it is 4200 J kg⁻¹ °C⁻¹. Find the final temperature θ, assuming no heat is lost.
- Heat lost by the metal = 0.50 × 400 × (100 − θ) = 200(100 − θ).
- Heat gained by the water = 0.20 × 4200 × (θ − 20) = 840(θ − 20).
- Set them equal: 20 000 − 200θ = 840θ − 16 800.
- Rearrange: 36 800 = 1040θ, so θ = 35°C (to 2 significant figures).
The final temperature lies between 20°C and 100°C, and it is closer to the water’s start because the water can absorb more heat per degree.
The mistake that costs marks
The slip is to say “cold flows into the hot object”, or to assume the final temperature is the average of the two starting temperatures. The average works only when the heat capacities are equal.
| Statement | Problem | Correct |
|---|---|---|
| “Cold flows into the hot object” | Heat is the thing that flows | Heat flows from hot to cold |
| θ = (100 + 20) ÷ 2 = 60°C | Ignores m and c | Use heat lost = heat gained |
| “Equilibrium means heat stops moving” | Particles still exchange energy | No net heat flow |
Check your answer for sense. The final temperature must fall between the two starting values. The units and significant figure checker helps you keep the units tidy.
Check yourself
0.10 kg of water at 90°C is mixed with 0.40 kg of water at 20°C. Find the final temperature, assuming no heat is lost.
Answer
Both samples are water, so c cancels. Heat lost: 0.10(90 − θ). Heat gained: 0.40(θ − 20).
Then 9 − 0.1θ = 0.4θ − 8, so 17 = 0.5θ and θ = 34°C.
The result is close to the cold water’s start, because there is four times as much cold water.
What to study next
Go on to using specific heat capacity, which gives the formula used in the mixing example. Then test yourself with the heat practice set.
To have a teacher check your explanations and mixing calculations, see online one-to-one Physics tuition.