# 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.

സംവേദനാത്മക പതിപ്പ്: https://www.calcopenly.com/ml/science/heat-transfer-calculator
വിഷയം: ശാസ്ത്ര കാൽക്കുലേറ്ററുകൾ

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.

## ഇൻപുട്ടുകൾ

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

## ഫലങ്ങൾ

- Heat (J) — പ്രധാന ഫലം
- പിണ്ഡം (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

## സൂത്രവാക്യം

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

## പരിഹരിച്ച ഉദാഹരണങ്ങൾ

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

- Calculate: Temperature change
- Solve for: Heat
- Material: Water (liquid, 15 °C)
- പിണ്ഡം: 1 kg
- Initial temperature: 20 °C
- Final temperature: 100 °C
- **Heat: 334,880 J**
- **Heat: 0.093022 kWh**
- പരിശോധനയുടെ ഉറവിടം: 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)
- പിണ്ഡം: 1 kg
- Initial temperature: 80 °C
- Final temperature: 20 °C
- **Heat: -251,160 J**
- പരിശോധനയുടെ ഉറവിടം: 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
- പിണ്ഡം: 500 g
- Initial temperature: 20 °C
- **Final temperature: 40 °C**
- **Temperature change: 20 K**
- പരിശോധനയുടെ ഉറവിടം: 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
- പിണ്ഡം: 1 kg
- Initial temperature: 20 °C
- Final temperature: 30 °C
- **Specific heat capacity: 387 J/(kg·K)**
- പരിശോധനയുടെ ഉറവിടം: Python 3.8: 3870/(1 × 10) = 387 J/(kg·K), copper in OpenStax Table 1.3

### Melt 2 kg of ice

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

### Boil away 500 g of water

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

## ചോദ്യങ്ങൾ

### 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” എത്രത്തോളം കൃത്യമാണ്?

കൃത്യത നിങ്ങളുടെ ഇൻപുട്ടുകളെയും രീതിയുടെ അനുമാനങ്ങളെയും ആശ്രയിച്ചിരിക്കുന്നു. ദശാംശ ഗണിതം 50 സാർഥക അക്കങ്ങൾ ഉപയോഗിക്കുന്നു. എന്നാൽ അനുമാനക്കണക്കുകൾ, സംഖ്യാത്മക രീതികൾ, ഉറവിട ഡാറ്റ എന്നിവയ്ക്ക് കൃത്യത കുറവാകാം; പ്രദർശിപ്പിക്കുന്ന മൂല്യം റൗണ്ട് ചെയ്യുന്നത് ഈ പരിമിതികൾ നീക്കില്ല. സ്വതന്ത്ര ഉറവിടങ്ങളിലെ പരിഹാരങ്ങളുമായി പരിശോധിച്ച ഉദാഹരണങ്ങൾ: 10. ഉദാഹരണത്തിന്, “Heat 1 kg of water from 20 °C to 100 °C” എന്നത് Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186) ഉപയോഗിച്ച് പരിശോധിക്കുന്നു.

### ഈ രീതിയുടെ ഉറവിടം എന്താണ്?

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.

## സ്രോതസ്സുകൾ

- [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)