# 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.

النسخة التفاعلية: https://www.calcopenly.com/ar/science/heat-transfer-calculator
الموضوع: حاسبات العلوم

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.

## المدخلات

- **Calculate** (الخيارات: Temperature change, Phase change, Carnot efficiency)
- **Solve for** (الخيارات: Heat, الكتلة, Final temperature, Specific heat)
- **Material** (الخيارات: Water (liquid, 15 °C), Ice (average −50 to 0 °C), Aluminium, Copper, Iron / steel, Lead, Silver, Gold, Glass, Concrete / granite, الخشب, Ethanol, Mercury, Human body (average), Enter specific heat)
- **Specific heat capacity**
- **Heat added (negative if removed)**
- **الكتلة**
- **Initial temperature**
- **Final temperature**
- **Solve for** (الخيارات: Heat, الكتلة)
- **Substance** (الخيارات: Water, Ethanol, Nitrogen, Lead, Enter latent heat)
- **Change** (الخيارات: Melting / freezing, Boiling / condensing)
- **Latent heat**
- **Show heat in** (الخيارات: J, kJ, kcal, BTU)
- **Show temperatures in** (الخيارات: °C, °F, K)
- **Hot reservoir temperature**
- **Cold reservoir temperature**
- **Heat taken from the hot side**

## النتائج

- Heat (J) — النتيجة الرئيسية
- الكتلة (kg)
- Final temperature (°C)
- Specific heat capacity (J/(kg·K))
- Temperature change (K)
- Heat (kWh)
- Carnot efficiency
- Maximum work from that heat (J)
- Best refrigerator COP
- Best heat-pump COP

## الصيغة

$$
Q = mc\Delta T,\qquad Q = mL,\qquad \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}
$$

## 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

### Melt 2 kg of ice

- Calculate: Phase change
- الكتلة: 2 kg
- Solve for: Heat
- Substance: Water
- Change: Melting / freezing
- **Heat: 668,000 J**
- مصدر التحقق: ⁨Python 3.8: 2 × 334 kJ/kg (OpenStax Table 1.4)⁩

### Boil away 500 g of water

- Calculate: Phase change
- الكتلة: 0.5 kg
- Solve for: Heat
- Substance: Water
- Change: Boiling / condensing
- **Heat: 1,128,000 J**
- مصدر التحقق: ⁨Python 3.8: 0.5 × 2256 kJ/kg (OpenStax Table 1.4)⁩

## 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.

## المصادر

- [OpenStax University Physics Volume 2, §1.5 Heat transfer, specific heat and calorimetry (Table 1.3); §1.6 Phase changes (Table 1.4)](https://openstax.org/books/university-physics-volume-2/pages/1-5-heat-transfer-specific-heat-and-calorimetry)
- [OpenStax University Physics Volume 2, §4.5 The Carnot cycle](https://openstax.org/books/university-physics-volume-2/pages/4-5-the-carnot-cycle)