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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.

آخر تحديث أمثلة تم التحقق منها: 10

خيارات إضافية
جرّب
Heat
J
Heat: 334,880 J
الأرقام المعنوية: 6؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر
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
طريقة الحساب 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}

حول 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.

أمثلة محلولة

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

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

مصدر التحقق: ⁨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)
الكتلة
1 kg
Initial temperature
80 °C
Final temperature
20 °C
Heat
-251,160 J

مصدر التحقق: ⁨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
الكتلة
500 g
Initial temperature
20 °C
Final temperature
40 °C
Temperature change
20 K

مصدر التحقق: ⁨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
الكتلة
1 kg
Initial temperature
20 °C
Final temperature
30 °C
Specific heat capacity
387 J/(kg·K)

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

الأسئلة

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.

ما مدى دقة «⁨Specific heat, latent heat and Carnot efficiency calculator⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 10. مثلًا، يجري التحقق من «⁨Heat 1 kg of water from 20 °C to 100 °C⁩» بالرجوع إلى ⁨Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186)⁩.

ما مصدر هذه الطريقة؟

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.

حول هذه الحاسبة

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

المصادر

  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

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