Temperature and Heat

Thermal Equilibrium

equal capacities meet halfway

When equal-capacity objects touch and settle, the final temperature is the average of the initial kelvin readings. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start with two thermal states

One object is at 300 kelvin. The other is at 360 kelvin. Give them equal heat capacity.

TA=300 KTB=360 KT_A = 300\ \text{K}\qquad T_B = 360\ \text{K}
Thermal equilibriumEqual-capacity states and the final state share one scale.300 Kcold360 Khot330 Kfinal

Equal capacities meet at the average

With equal heat capacity, the warmer object loses the same heat the cooler object gains. The final temperature is halfway.

Tf=TA+TB2T_f = \frac{T_A + T_B}{2}

Compute the final temperature

Add 300 kelvin and 360 kelvin, then divide by 2. The shared final state is 330 kelvin.

Tf=300 K+360 K2=330 KT_f = \frac{300\ \text{K} + 360\ \text{K}}{2} = 330\ \text{K}
Thermal equilibriumEqual-capacity states and the final state share one scale.300 Kcold360 Khot330 Kfinal

Equal capacities meet at the midpoint

The diagram shows the first row. Equal capacities make the final temperature the midpoint between the two initial kelvin readings.

TATBTf300 K360 K330 K310 K350 K330 K300 K340 K320 K\begin{array}{c|c|c}T_A&T_B&T_f\\300\ \text{K}&360\ \text{K}&330\ \text{K}\\310\ \text{K}&350\ \text{K}&330\ \text{K}\\300\ \text{K}&340\ \text{K}&320\ \text{K}\\\end{array}
Thermal equilibriumThe first table row is the checked diagram.300 Kcold360 Khot330 Kfinal