For a fixed material, heat is directly proportional to both mass and temperature change.

highlighted = computed this step

Start from the base case

The base case needs 30 joules of heat.

Qbase=30 JQ_{\text{base}} = 30\ \text{J}
Heat contrastComparable bars share one energy scale.30 Jbase60 Jmore mass60 Jmore warming

Double the mass

Keeping the material and warming the same, doubling mass doubles the heat to 60 joules.

Qmore mass=230 J=60 JQ_{\text{more mass}} = 2\cdot 30\ \text{J} = 60\ \text{J}

Double the temperature change

Keeping the mass and material the same, doubling the temperature change also gives 60 joules.

Qmore warming=230 J=60 JQ_{\text{more warming}} = 2\cdot 30\ \text{J} = 60\ \text{J}
Heat contrastComparable bars share one energy scale.30 Jbase60 Jmore mass60 Jmore warming