NapkinCalc

Physics 4 — Thermo & Fluids

Heat & specific heat

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

ELI5: heat is energy in transit, and warming something costs Q = m·c·ΔT. The "c" is how stubborn a material is about changing temperature — and water's is enormous (4186 J/kg·K), which is why the oceans tame the climate and why boiling a kettle takes real time.

mw:=1kgm_{w} := 1 kg = 1 kg one liter of water
cw:=4186J/(kg∗K)c_{w} := 4186 J/(kg*K) = 4186 J / (kg K) specific heat of water
dT:=80KdT := 80 K = 80 K from 20 °C to boiling
Qboil=mw⋅cw⋅dTQ_{boil} = m_{w} \cdot c_{w} \cdot dT = 334880 J ~335 kJ
tkettle:=Qboil/(2000W)t_{kettle} := Q_{boil} / (2000 W) = 167.44 s a 2 kW kettle needs ~167 s — time yours!

Real-world hook: high specific heat is why coastal cities have mild weather, why a car radiator uses water, and why the filling of an apple pie burns your mouth long after the crust has cooled.

Try it yourself: how much heat (J) warms 2 kg of water by 30 K? (Q = m·c·ΔT, with c = c_w above.)

Qyou:=2kg∗cw∗30KQ_{you} := 2 kg * c_{w} * 30 K = 251160 J ✏️ Your turn: multiply 2 kg · c_w · 30 K. Keep the unit (joules).
✓ pass abs(Qyou−2kg∗cw∗30K)<1Jabs(Q_{you} - 2 kg * c_{w} * 30 K) < 1 J green when your heat is correct