Electricity Consumption of a TV
Do you remember when TVs used flickering cathode‑ray tubes? Screens were much smaller back then — typically around 20‑inches instead of today’s 65‑inch models. Which matters more for energy use: improvements in efficiency, or the increase in screen area?
| Technology | Power per Screen Area |
|---|---|
| Good Old Tube | 500 W/m2 |
| Plasma | 400 W/m2 |
| LED | 120 W/m2 |
| OLED | 180 W/m2 |
Calculate the energy consumption and yearly bill by changing the inputs below. For an old TV, use a 20‑inch CRT and 500 W/m²; for a modern LED TV, use a 65‑inch screen and 120 W/m². Set the aspect ratio to 4:3 for CRTs and 16:9 for modern displays.
And how many hours a day is your TV running?
| Screen Geometry | |
| =============== | |
| Typically, we know the diagonal and the aspect ratio. | |
| Diagonal: d = 20 inch | 20 inch |
| Aspect ratio: ar = 4 / 3 | 1.33333 |
| How to get height, width and area? | |
| Pythagoras: d = sqrt(w^2 + h^2) | |
| Divided by h: d/h = sqrt((w/h)^2 + 1^2) | |
| Using aspect ratio: d/h = sqrt(ar^2+1) | |
| We can solve for h and finally calculate: | |
| Height: h = d / sqrt(1 + ar^2) | 12 inch |
| Width: w = h * ar | 16 inch |
| Area: A = h * w to m2 | 0.123871 m2 |
| Yearly Energy Bill | |
| ================== | |
| Power per Area: ppA = 500 W/m2 | 500 W/m2 |
| Power: P = ppA * A | 61.9354 W |
| Energy: E = 5 h/day * 1 year * P to kWh | 113.109 kWh |
| Cost: c = 14 USDct/kWh * E to $ | 15.8353 $ |
In case you already know the power your TV needs, it is a one-liner:
| TV energy bill: 150 W * 6 h/day * 1 year * 37 EURct/kWh | 121.628 € |
Can You Cook Pasta on Mount Everest?
How does the low air pressure affect pasta cooking?
In high altitude, the air pressure is much lower. As a consequence, water will boil earlier than usual!
| Mountain height: | |
| h = 8849 m | 8849 m |
| Air pressure decreases by about 1% every 80 m: | |
| p = 1013 mbar * 99% ^ (h / 80 m) to mbar | 333.279 mbar |
| What is the boiling temperature of water? | |
| load fluid water | 1 |
| T_boil = T(p) | 71.5326 °C |
It is not the bubbles of the boiling water that get your pasta done, it is the temperature. And with only 72°C, you'd have to wait forever.
What about other Mountains, or the elevation of your home?
Edit the calculation above and change the height value
(Matterhorn: 4478 m, Kilimanjaro 5895, Mount Fuji 3776 m, Denali 6190 m, Dead Sea: -430 m).
Fluid Properties Calculator
2026-06-27 by Samuel
Define your fluid and use any pair of pressure, temperature, specific enthalpy, specific entropy, density and vapor quality as parameter 1 and 2. Calcumber then calculates all properties for that state.
See pure fluids and fluid mixtures for a list of available fluids.
| Fluid Selection: | |
| load fluid water | 1 |
| State parameters: | |
| Parameter1: par1 = 20 °C | 20 °C |
| Parameter2: par2 = 1 bar | 1 bar |
| Temperature: T(par1, par2) | 293.15 K |
| Pressure: p(par1, par2) | 1 bar |
| Specific enthalpy: h(par1, par2) | 84.006054 kJ/kg |
| Gibbs energy: g(par1, par2) | -2.9021070 kJ/kg |
| Inernal energy: u(par1, par2) | 83.905874 kJ/kg |
| Specific entropy: s(par1, par2) | 0.29646311 kJ/(kg*K) |
| Density: rho(par1, par2) | 998.20654 kg/m3 |
| Specivic volume: v(par1, par2) | 0.0010017967 m3/kg |
| Vapor quality: x(par1, par2) to % | -100 % |
| Speed of sound: a(par1, par2) | 1482.3440 m/s |
| Viscosity: mu(par1, par2) to mPa*s | 1.0015966 mPa*s |
| Thermal conductivity: k(par1, par2) | 0.59801157 W/(m*K) |
| Specific heat | |
| - at constant pressure: cp(par1, par2) | 4184.0551 J/(kg*K) |
| - at constant volume: cv(par1, par2) | 4156.6862 J/(kg*K) |