What quantity of heat must be removed from 20g of water at 0 OC to change it to ice at 0 OC?

This specific heat calculator is a tool that determines the heat capacity of a heated or a cooled sample. Specific heat is the amount of thermal energy you need to supply to a sample weighing 1 kg to increase its temperature by 1 K. Read on to learn how to apply the heat capacity formula correctly to obtain a valid result.

💡 This calculator works in various ways, so you can also use it to, for example, calculate the heat needed to cause a temperature change (if you know the specific heat). If you have to achieve the temperature change in a determined time, use our watts to heat calculator to know the power required. To find specific heat from a complex experiment, calorimetry calculator might make the calculations much faster.

Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you:

1. Determine whether you want to warm up the sample (give it some thermal energy) or cool it down (take some thermal energy away).
2. Insert the amount of energy supplied as a positive value. If you want to cool down the sample, insert the subtracted energy as a negative value. For example, say that we want to reduce the sample's thermal energy by 63,000 J. Then Q = -63,000 J.
3. Decide the temperature difference between the initial and final state of the sample and type it into the heat capacity calculator. If the sample is cooled down, the difference will be negative, and if warmed up - positive. Let's say we want to cool the sample down by 3 degrees. Then ΔT = -3 K. You can also go to advanced mode to type the initial and final values of temperature manually.
4. Determine the mass of the sample. We will assume m = 5 kg.
5. Calculate specific heat as c = Q / (mΔT). In our example, it will be equal to c = -63,000 J / (5 kg * -3 K) = 4,200 J/(kg·K). This is the typical heat capacity of water.

If you have problems with the units, feel free to use our temperature conversion or weight conversion calculators.

The formula for specific heat looks like this:

Q is the amount of supplied or subtracted heat (in joules), m is the mass of the sample, and ΔT is the difference between the initial and final temperatures. Heat capacity is measured in J/(kg·K).

You don't need to use the heat capacity calculator for most common substances. The values of specific heat for some of the most popular ones are listed below.

• ice: 2,100 J/(kg·K)
• water: 4,200 J/(kg·K)
• water vapor: 2,000 J/(kg·K)
• basalt: 840 J/(kg·K)
• granite: 790 J/(kg·K)
• aluminum: 890 J/(kg·K)
• iron: 450 J/(kg·K)
• copper: 380 J/(kg·K)
• lead: 130 J/(kg·K)

Having this information, you can also calculate how much energy you need to supply to a sample to increase or decrease its temperature. For instance, you can check how much heat you need to bring a pot of water to the boil to cook some pasta.

Wondering what the result actually means? Try our potential energy calculator to check how high you would raise the sample with this amount of energy. Or check how fast could the sample move with this kinetic energy calculator.

1. Find the initial and final temperature as well as the mass of the sample and energy supplied.
2. Subtract the final and initial temperature to get the change in temperature (ΔT).
3. Multiply the change in temperature with the mass of the sample.
4. Divide the heat supplied/energy with the product.
5. The formula is C = Q / (ΔT ⨉ m).

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m).

The formula for specific heat capacity, C, of a substance with mass m, is C = Q /(m ⨉ ΔT). Where Q is the energy added and ΔT is the change in temperature. The specific heat capacity during different processes, such as constant volume, Cv and constant pressure, Cp, are related to each other by the specific heat ratio, ɣ= Cp/Cv, or the gas constant R = Cp - Cv.

Specific heat capacity is measured in J/kg K or J/kg C, as it is the heat or energy required during a constant volume process to change the temperature of a substance of unit mass by 1 °C or 1 °K.

The specific heat of water is 4179 J/kg K, the amount of heat required to raise the temperature of 1 g of water by 1 Kelvin.

Specific heat is measured in BTU / lb °F in imperial units and in J/kg K in SI units.

The specific heat of copper is 385 J/kg K. You can use this value to estimate the energy required to heat a 100 g of copper by 5 °C, i.e., Q = m x Cp x ΔT = 0.1 * 385 * 5 = 192.5 J.

The specific heat of aluminum is 897 J/kg K. This value is almost 2.3 times of the specific heat of copper. You can use this value to estimate the energy required to heat a 500 g of aluminum by 5 °C, i.e., Q = m x Cp x ΔT = 0.5 * 897* 5 = 2242.5 J.

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Heat of Fusion-the amount of heat required to convert unit mass of a solid into the liquid without a change in temperature. (or released for freezing)

For water at its normal freezing point of 0 ºC, the specific heat of Fusion is 334 J g-1. This means that to convert 1 g of ice at 0 ºC to 1 g of  water at 0 ºC, 334 J of heat must be absorbed by the water. Conversely, when 1 g of water at 0 ºC freezes to give 1 g of ice  at 0 ºC, 334 J of heat will be released to the surroundings.

Heat of Fusion of Water (Hf = 334 J /g)

q= m Hf

Note- The Heat of Fusion equation is used only at the melting/freezing transition, where the temperature remains the same only and that is why there is no temperature change (DT) in this formula. It stays at 0 Celsius for water.

Note #2-Energy is required to melt and released when it freezes

The diagram on the left shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at temperatures above 100 ºC, affects the temperature of the sample. E: Steam absorbs heat and thus increases its temperature. D: Water boils and absorbs latent heat of vaporization. C: Rise in temperature as liquid water absorbs heat. B: Absorption of latent heat of fusion. A: Rise in temperature as ice absorbs heat. from-http://www.physchem.co.za/Heat/Latent.htm

This worked example problem demonstrates how to calculate the energy required to raise the temperature of a sample that includes changes in phase. This problem finds the energy required to turn cold ice into hot steam.

What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?Useful information:heat of fusion of water = 334 J/gheat of vaporization of water = 2257 J/gspecific heat of ice = 2.09 J/g·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°C

The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up.

Step 1:

Find the heat required to raise the temperature of ice from -10 °C to 0 °C. Use the formula:

q = mcΔT

where

• q = heat energy
• m = mass
• c = specific heat
• ΔT = change in temperature

In this problem:

• q = ?
• m = 25 g
• c = (2.09 J/g·°C
• ΔT = 0 °C - -10 °C (Remember, when you subtract a negative number, it is the same as adding a positive number.)

Plug in the values and solve for q:

q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]q = (25 g)x(2.09 J/g·°C)x(10 °C)

q = 522.5 J

The heat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 J

Step 2:

Find the heat required to convert 0 °C ice to 0 °C water.

Use the formula for heat:

q = m·ΔHf

where

• q = heat energy
• m = mass
• ΔHf = heat of fusion

For this problem:

• q = ?
• m = 25 g
• ΔHf = 334 J/g

Plugging in the values gives the value for q:

q = (25 g)x(334 J/g)
q = 8350 J

The heat required to convert 0 °C ice to 0 °C water = 8350 J

Step 3:

Find the heat required to raise the temperature of 0 °C water to 100 °C water.q = mcΔTq = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]q = (25 g)x(4.18 J/g·°C)x(100 °C)q = 10450 JThe heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J

Step 4:

Find the heat required to convert 100 °C water to 100 °C steam.
q = m·ΔHvwhere

q = heat energy

ΔHv = heat of vaporization

The heat required to convert 100 °C water to 100 °C steam = 56425

Step 5:

Find the heat required to convert 100 °C steam to 150 °C steamq = mcΔTq = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]q = (25 g)x(2.09 J/g·°C)x(50 °C)q = 2612.5 J

The heat required to convert 100 °C steam to 150 °C steam = 2612.5

Step 6:

Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.

HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5
HeatTotal = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J
HeatTotal = 78360 J

Answer:

The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.

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• Ge, Xinlei; Wang, Xidong (2009). "Calculations of Freezing Point Depression, Boiling Point Elevation, Vapor Pressure and Enthalpies of Vaporization of Electrolyte Solutions by a Modified Three-Characteristic Parameter Correlation Model". Journal of Solution Chemistry. 38 (9): 1097–1117. doi:10.1007/s10953-009-9433-0
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