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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.

અપડેટ કરેલું તપાસેલાં ઉદાહરણો: 10

વધુ વિકલ્પો
અજમાવો
Heat
J
Heat: 334,880 J
સૂચક અંકો: 6; સૌથી નજીક; સરખા અંતરે શૂન્યથી દૂર
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
ગણતરી કેવી રીતે થાય છે 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 વિશે

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.

ઉકેલેલાં ઉદાહરણો

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

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

ચકાસણીનો સ્ત્રોત: 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)
દળ
1 kg
Initial temperature
80 °C
Final temperature
20 °C
Heat
-251,160 J

ચકાસણીનો સ્ત્રોત: 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
દળ
500 g
Initial temperature
20 °C
Final temperature
40 °C
Temperature change
20 K

ચકાસણીનો સ્ત્રોત: 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
દળ
1 kg
Initial temperature
20 °C
Final temperature
30 °C
Specific heat capacity
387 J/(kg·K)

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

પ્રશ્નો

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” કેટલું ચોક્કસ છે?

ચોકસાઈ તમારા ઇનપુટ અને પદ્ધતિની ધારણાઓ પર આધારિત છે. દશાંશ ગણતરી 50 સાર્થક અંકો વાપરે છે, પરંતુ અંદાજ, સંખ્યાત્મક પદ્ધતિઓ અને મૂળ ડેટા ઓછા ચોક્કસ હોઈ શકે છે; દર્શાવેલા મૂલ્યોને રાઉન્ડ કરવાથી આ મર્યાદાઓ દૂર થતી નથી. સ્વતંત્ર સ્ત્રોતોના ઉકેલેલા ઉદાહરણો સાથે ચકાસણી: 10. ઉદાહરણ તરીકે, “Heat 1 kg of water from 20 °C to 100 °C”ને Python 3.8 fractions: 1 × 4186 × 80 = 334880 J (OpenStax c_water = 4186) સાથે ચકાસવામાં આવે છે.

આ પદ્ધતિનો સ્ત્રોત શું છે?

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.

આ કેલ્ક્યુલેટર વિશે

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

સ્રોતો

  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

સંદર્ભો સાથે ચકાસેલાં

આ કેલ્ક્યુલેટરમાં સ્વતંત્ર સ્રોતોના જવાબો ધરાવતા 10 ઉકેલેલાં ઉદાહરણો છે. તે પરીક્ષણ સમૂહમાં ચાલે છે અને તમે અહીં પણ ચલાવી શકો છો.

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