# Which one will require the most energy to raise the temperature of a 50 g sample from 20 c to 30 c

1) Calculate the energy required to boil 100ml of water for a cup of tea if the initial water temperature is 27.0°C. (The density of water is 1g/ml)

Since the density of water is 0.997g/ml , at 25oC we can round it to 1.00 g/mL. So 100ml of water has a mass of 100 grams. The change in temperature is (100°C - 27°C) = 73°C. Since the specific heat of water is 4.18J/g/°C we can calculate the amount of energy needed by the expression below.

Energy required = 4.18 J/g/°C X 100g X 73°C = 30.514KJ.

Try some exercises. 1) Calculate the energy needed to heat a) 80ml of water form 17°C to 50°C; b) 2.3 litres of water from 34°C to 100°C; c) 200g of cooking oil from 23°C to 100°C

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An astronaut in space needs to absorb 2,400KJ of solar energy in a container with an accurately known volume of water. The water's temperature is needed to increase from 20°C to 34.5°C. What amount of water is in the container?
Solution

3,450,560J of energy are absorbed by 300kg of water. If the initial temperature of the water is 20°C what is the final temperature?
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A peanut of mass 2.34g is burnt in a calorimeter containing 100ml of water. If the temperature of the calorimeter rises from 23.5°C to 27.7°C calculate the energy content of the peanut in joules per gram.
Solution

When 0.15 gram of heptane C7H16 was burnt in a bomb calorimeter containing 1.5kg of water the temperature rose from 22.000°C to 23.155°C. Calculate the heat given out by heptane during combustion per mole. This is known as the heat of combustion.
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0.1 gram of methane was completely burnt in a bomb calorimeter containing 100ml of water. If the temperature increased by 11.82°C find the heat of combustion(energy released per mole)of methane.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
4.94KJ = 4.18J/g/°C X 11.82°C X 100
moles of methane = 0.1/16 = 0.00625mole
4.94 / 0.00625 = 790.4KJ/mole

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A brand of RAZMAN'S jelly beans was analysed for its energy content. A jelly bean of mass 1.5g was burnt completey in a bomb calorimeter containing 100ml of water. If the temperature rose from 21.56°C to 24.5°C calculate the energy content per gram of the jelly bean.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
1.23KJ = 4.18J/g/°C X 2.94°C X 100g
1.23/1.5 = 0.819KJ/g

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0.1 gram of carbon was burnt in a bomb calorimeter containing 200ml of water. If the temperature of the water increased by 3.92°C calculate the energy given off, during combustion, per mole of carbon.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
3277.12J = 4.18J/g/°C X 3.92°C X 200
Mole of carbon = 0.1 / 12 =0.0083

Energy per mole = 3277.12J / 0.0083 =394,819J/mole

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A 2.6 gram sample of sugar was burnt in a bomb calorimeter containing 100ml of water. The temperature of the water increased from 22.0°C to 24.3°C.

a)Calculate the amount of energy that was released in the burning of the sugar.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
961.4J = 4.18J/g/°C X 2.3°C X 100

b) Calculate the mole and the mass of sugar present.

Assuming all the energy released is absorbed by the water,every mole of sugar burnt must release 2803KJ of energy. The amount of sugar present in mole is given by the expression below 0.9614KJ/2803KJ = 0.00034mole

Mass of sugar = 0.00034 X 180 = 0.0612 grams

c) Assuming no other combustible material is present calculate the percent, by mass, of sugar in the sample.

(0.0612 / 2.6 ) X 100 = 2.35%

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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:

c = Q / (mΔT)

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)

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.