A shared start temperature and three checked endpoints show how Delta T controls heat. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Hold mass and material fixed

Now keep the sample mass at 2 kilograms and the specific heat at 5 joules per kilogram per kelvin.

Q=mcΔTQ = mc\Delta T
Temperature-change scanThe thermometer endpoints and heat bars are checked together.20 Jheat40 Jheat60 Jheat300 Kstart302 Kend304 Kend306 Kend

First warming point

The first endpoint is only a small temperature change from the same start.

Qfirst=2 kg5 J/(kg K)2 K=20 JQ_{\text{first}} = 2\ \text{kg}\cdot 5\ \text{J/(kg K)}\cdot 2\ \text{K} = 20\ \text{J}

Second warming point

The middle endpoint doubles the temperature change while mass and material stay fixed.

Qsecond=2 kg5 J/(kg K)4 K=40 JQ_{\text{second}} = 2\ \text{kg}\cdot 5\ \text{J/(kg K)}\cdot 4\ \text{K} = 40\ \text{J}

Third warming point

The third endpoint gives a third point on the same direct relationship.

Qthird=2 kg5 J/(kg K)6 K=60 JQ_{\text{third}} = 2\ \text{kg}\cdot 5\ \text{J/(kg K)}\cdot 6\ \text{K} = 60\ \text{J}

Read the three-point temperature scan

The thermometer endpoints move farther from the common start, and the heat bars grow on the same scale.

mcΔTQ2 kg5 J/(kg K)2 K20 J2 kg5 J/(kg K)4 K40 J2 kg5 J/(kg K)6 K60 J\begin{array}{c|c|c|c}m&c&\Delta T&Q\\2\ \text{kg}&5\ \text{J/(kg K)}&2\ \text{K}&20\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&4\ \text{K}&40\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&6\ \text{K}&60\ \text{J}\\\end{array}
Temperature-change scanThe thermometer endpoints and heat bars are checked together.20 Jheat40 Jheat60 Jheat300 Kstart302 Kend304 Kend306 Kend