Heat energy from Q = mcΔT (solve for heat, mass, final temperature or specific heat), latent heat Q = mL for melting or boiling, and Carnot efficiency.
Aggiornato Esempi verificati: 10
Adding 334.88 kJ to 1 kg of material with c = 4,186 J/(kg·K) changes its temperature by 80 K, from 20 °C to 100 °C.
A change of 1 °C equals a change of 1 K.
Warming or cooling a material takes heat Q = mcΔT, where m is the mass, c the specific heat capacity and ΔT the temperature change; the calculator solves for any one of heat, mass, final temperature or specific heat. Melting or boiling takes Q = mL at constant temperature, where L is the latent heat. The Carnot mode gives the upper limit on any heat engine's efficiency, 1 − Tc/Th with both temperatures in kelvin.
The default, 1 kg of water heated from 20 °C to 100 °C, needs 334,880 J (0.093 kWh), what a 2 kW kettle delivers in 2 minutes 47 seconds with no losses. Boiling that water away takes a further 2,256 kJ, almost seven times as much.
Specific heats are room-temperature values from OpenStax University Physics (Table 1.3). In reality c varies with temperature, and the calculation assumes no melting or boiling between the two temperatures.
Fonte di verifica: Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186)
Fonte di verifica: Python 3.8: 1 × 4186 × (20 − 80)
Fonte di verifica: Python 3.8 fractions: ΔT = 9000/(0.5 × 900) = 20 K
Fonte di verifica: Python 3.8: 3870/(1 × 10) = 387 J/(kg·K), copper in OpenStax Table 1.3
4,186 J per kilogram per degree Celsius, water's specific heat capacity. Heating 1 litre (1 kg) from 20 °C to 100 °C takes 1 × 4,186 × 80 = 334,880 J, or 0.093 kWh. In US units that is about 1 BTU per pound per °F, which is how the BTU was originally defined.
The heat needed to raise 1 kg of a substance by 1 K (the same as 1 °C), in J/(kg·K). Water's is 4,186, among the highest of common substances, while copper's is 387 and lead's 128. The same 10 kJ warms 1 kg of water by 2.4 °C but 1 kg of copper by 25.8 °C, which is why water is used for cooling and heat storage.
The energy absorbed or released during a phase change at constant temperature, Q = mL. For water the latent heat of fusion is 334 kJ/kg at 0 °C and of vaporisation 2,256 kJ/kg at 100 °C (OpenStax Table 1.4). Melting 2 kg of ice takes 668 kJ, enough to heat the same 2 kg of water by about 80 °C.
η = 1 − Tc/Th, the largest fraction of heat that any engine can turn into work between a hot reservoir at Th and a cold one at Tc, both in kelvin. Between 500 K and 300 K it is 40%; between boiling and freezing water, 26.8%. Real engines fall short of it: coal-fired power stations typically convert about 37% of their fuel's heat into electricity.
COP = Th/(Th − Tc) with temperatures in kelvin, the Carnot limit on heat delivered per unit of work. Pumping heat from 0 °C outdoors into a 35 °C heating loop allows at most 308.15/35 ≈ 8.8. At −10 °C outside the limit drops to 6.8, and real machines stay well below it because of compressor and heat-exchanger losses.
La precisione dipende dai dati inseriti e dalle ipotesi del metodo. Il calcolo decimale usa 50 cifre significative, ma stime, metodi numerici e dati di origine possono essere meno precisi; l’arrotondamento visualizzato non elimina questi limiti. Esempi svolti verificati con fonti indipendenti: 10. Per esempio, «Heat 1 kg of water from 20 °C to 100 °C» viene verificato con Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186).
OpenStax University Physics Volume 2, §1.5 Heat transfer, specific heat and calorimetry (Table 1.3); §1.6 Phase changes (Table 1.4); OpenStax University Physics Volume 2, §4.5 The Carnot cycle.
Questa calcolatrice include 10 esempi svolti con risposte da fonti indipendenti. Fanno parte della suite di test e puoi eseguirli anche qui.
Solve the ideal gas law PV = nRT for pressure, volume, moles or temperature in any units, or find a new state with the combined gas law.
Kinetic energy (½mv²), potential energy (mgh), spring energy, work (Fd cos θ) and power (W/t), solved for any variable, plus the speed after a drop.
Electricity cost of running an appliance per hour, day, month and year, from its wattage, hours of use and your price per kWh (unit).