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.

Güncellendi Doğrulanan örnekler: 10

Diğer seçenekler
Dene
Heat
J
Heat: 334,880 J
Anlamlı basamaklar: 6; En yakına; eşit uzaklıkta sıfırdan uzağa
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
Nasıl hesaplanır 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}

Specific heat, latent heat and Carnot efficiency calculator hakkında

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.

Çözümlü örnekler

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

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

Doğrulama kaynağı: 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)
Kütle
1 kg
Initial temperature
80 °C
Final temperature
20 °C
Heat
-251,160 J

Doğrulama kaynağı: 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
Kütle
500 g
Initial temperature
20 °C
Final temperature
40 °C
Temperature change
20 K

Doğrulama kaynağı: 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
Kütle
1 kg
Initial temperature
20 °C
Final temperature
30 °C
Specific heat capacity
387 J/(kg·K)

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

Sorular

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.

“Specific heat, latent heat and Carnot efficiency calculator” ne kadar doğru sonuç verir?

Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 10. Örneğin “Heat 1 kg of water from 20 °C to 100 °C”, Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186) ile karşılaştırılarak doğrulanır.

Yöntemin kaynağı nedir?

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.

Bu hesaplayıcı hakkında

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

Kaynaklar

  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

Kaynaklarla doğrulandı

Bu hesaplayıcı, yanıtları bağımsız kaynaklardan alınan 10 çözümlü örnek içerir. Bunlar test paketinde çalıştırılır; burada da çalıştırabilirsiniz.

İlgili hesaplayıcılar