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Specific heat, latent heat and Carnot efficiency calculator

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.

Aktualisiert Geprüfte Beispiele: 10

Weitere Optionen
Ausprobieren
Heat
J
Heat: 334,880 J
Signifikante Stellen: 6; Zum nächsten Wert, bei Gleichstand von null weg
Specific heat capacity
4,186J/(kg·K)
Temperature change
80K
Heat
0.0930222kWh
Assumes c stays constant over the range and no melting or boiling happens in between.

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.

Temperature against heat added

02550751000100200300Heat added (kJ)Temperature (°C)100 °C
So wird gerechnet S

  1. Temperature change

    ΔT=T2−T1=100−(20)=80 K\Delta T = T_2 - T_1 = 100 - (20) = 80\ \mathrm{K}

    A change of 1 °C equals a change of 1 K.

  2. Heat

    Q=mcΔT=(1)(4,186)(100−20)=334,880 JQ = mc\Delta T = (1)(4{,}186)(100 - 20) = 334{,}880\ \mathrm{J}

Über Specific heat, latent heat and Carnot efficiency calculator

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.

Durchgerechnete Beispiele

Heat 1 kg of water from 20 °C to 100 °C

Calculate
Temperature change
Solve for
Heat
Material
Water (liquid, 15 °C)
Masse
1 kg
Initial temperature
20 °C
Final temperature
100 °C
Heat
334,880 J
Heat
0.093022 kWh

Prüfquelle: Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186)

Cooling releases heat (negative Q)

Calculate
Temperature change
Solve for
Heat
Material
Water (liquid, 15 °C)
Masse
1 kg
Initial temperature
80 °C
Final temperature
20 °C
Heat
-251,160 J

Prüfquelle: Python 3.8: 1 × 4186 × (20 − 80)

9 kJ into 500 g of aluminium at 20 °C

Calculate
Temperature change
Solve for
Final temperature
Material
Aluminium
Heat added (negative if removed)
9 kJ
Masse
500 g
Initial temperature
20 °C
Final temperature
40 °C
Temperature change
20 K

Prüfquelle: Python 3.8 fractions: ΔT = 9000/(0.5 × 900) = 20 K

Identify a metal: 3870 J warms 1 kg by 10 K

Calculate
Temperature change
Solve for
Specific heat
Heat added (negative if removed)
3870 J
Masse
1 kg
Initial temperature
20 °C
Final temperature
30 °C
Specific heat capacity
387 J/(kg·K)

Prüfquelle: Python 3.8: 3870/(1 × 10) = 387 J/(kg·K), copper in OpenStax Table 1.3

Fragen

How much energy does it take to heat water?

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.

What is specific heat capacity?

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.

What is latent heat?

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.

What is the Carnot efficiency?

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.

What is the maximum COP of a heat pump?

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.

Wie genau arbeitet „Specific heat, latent heat and Carnot efficiency calculator“?

Die Genauigkeit hängt von Ihren Eingaben und den Annahmen der Methode ab. Die Dezimalrechnung nutzt 50 signifikante Stellen, doch Schätzungen, numerische Verfahren und Quelldaten können ungenauer sein. Die angezeigte Rundung beseitigt diese Grenzen nicht. Anhand unabhängiger Quellen geprüfte Rechenbeispiele: 10. Beispielsweise wird „Heat 1 kg of water from 20 °C to 100 °C“ anhand von Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186) geprüft.

Woher stammt die Methode?

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.

Über diesen Rechner

Q=mcΔT,Q=mL,ηCarnot=1−TCTHQ = mc\Delta T,\qquad Q = mL,\qquad \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Quellen

  1. OpenStax University Physics Volume 2, §1.5 Heat transfer, specific heat and calorimetry (Table 1.3); §1.6 Phase changes (Table 1.4)
  2. OpenStax University Physics Volume 2, §4.5 The Carnot cycle

Anhand von Quellen geprüft

Dieser Rechner enthält 10 Rechenbeispiele mit Ergebnissen aus unabhängigen Quellen. Sie laufen in der Testsuite und können auch hier ausgeführt werden.

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