NapkinCalc

Thermodynamics — Energy & Cycles

Specific heat — how much energy to warm something

continues from lesson 1 — values defined earlier in the course stay live here

ELI5: water is stubborn to heat; metal is easy. Specific heat c is how many joules raise one kilogram by one kelvin, and Q = m·c·ΔT is the total. Water's huge c (4186 J/kg·K) is why oceans moderate climate and why coolant works.

mwater:=2kgm_{water} := 2 kg = 2 kg a pot of water
cwater:=4186J/(kg∗K)c_{water} := 4186 J/(kg*K) = 4186 J / (kg K) water's specific heat
dTwarm:=30KdT_{warm} := 30 K = 30 K desired temperature rise
Qheat:=mwater∗cwater∗dTwarminkJQ_{heat} := m_{water} * c_{water} * dT_{warm} in kJ = 251.16 kJ energy needed ≈ 251 kJ
✓ pass abs(Qheat−251.16kJ)<0.1kJabs(Q_{heat} - 251.16 kJ) < 0.1 kJ a lot of energy — water resists warming

Try it yourself: the heat to warm 0.5 kg of water by 40 K? (Q = m·c·ΔT, c = 4186 J/(kg·K); answer in kJ.)

Qheatyou:=0.5kg∗4186J/(kg∗K)∗40KQheat_{you} := 0.5 kg * 4186 J/(kg*K) * 40 K = 83720 J ✏️ Your turn: multiply mass × specific heat × temperature rise.
✓ pass abs(Qheatyou−0.5kg∗4186J/(kg∗K)∗40K)<1e−6kJabs(Qheat_{you} - 0.5 kg * 4186 J/(kg*K) * 40 K) < 1e-6 kJ green when your heat input is correct