What is defined as the amount of heat energy required to raise the temperature of 1 gram of a substance by 1 degree 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

Solutions

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?
Solution

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.
Solution

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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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Hello, and welcome to this Mometrix video on specific heat capacity—a constant that relates heat transfer to changes in temperature.

Temperature is directly related to the average translational kinetic energy of the atoms or molecules in a system. Basically, the faster and heavier the particles are, the higher the temperature. The units for temperature are degrees Celsius and Kelvin (remember, 0 degrees Celsius = 273 Kelvin, but 1 degree Celsius has the same magnitude as 1 Kelvin).

Conversely, heat is measured in joules and is the energy transferred between systems at different temperatures that are in contact. Because heat is the transfer of energy, it is known as a process quantity.

When heat is absorbed or released by a system, the temperature changes. How much the temperature changes depends on the substance, and specifically, the specific heat capacity of that substance.

Let’s look at an example.

Let’s say we have 47.8 grams of water at 35ºC and we put it on the stove. We turn on the stove and transfer 1,000 joules of heat to our water and the temperature rises to 40ºC. In other words, it took 1,000 joules of heat to raise 47.8 grams of water by 5ºC.

That’s kind of a mouthful and seems oddly specific in terms of quantities. This is where specific heat capacity, notated as \(c\), comes into play. It’s a standard quantity and is the amount of heat required to raise 1 gram of a substance by 1 degree Celsius.

With some simple division, we can derive the specific heat capacity of water from our hypothetical cup of water.

\(\text{Specific heat capacity (}c\text{)} =\) \(qmass \times ΔT= \frac{1000\text{ joules}}{47.8\text{ g} \times 5\text{ K}}\)\(=4.184\text{ J/gK}\)

From this, we know now that it takes 4.184 joules to raise the temperature of 1 gram of water by 1 degree Celsius—that’s the specific heat capacity of water.

This is really helpful to scientists because once determined, the specific heat capacity can be used to calculate the heat absorbed or released by a system simply by measuring the temperature change and mass. Let’s try that out.

Let’s look at a new cup of water, let’s say a mug of 350 grams that’s boiling. The water starts at 100ºC and cools down to 90ºC. We want to know how much heat was released from the water to the surrounding environment. Since we know the specific heat capacity of water is 4.184 joules per gram Kelvin, we simply need to rearrange our previous equation to solve for \(q\) (the heat released).

\(q=c \times \text{mass} \times ∆T\)\(=4.184 \text{ J/gK} \times 350 \text{g} \times -10 \text{ K}\)\(=-14,644 \text{ J}\)

Since we knew the specific heat capacity of water, calculating the heat released from the system was easy!

Note here that the negative sign simply tells us that the system (the mug of water) released heat to the surrounding system rather than absorbed it.

Now that we’ve defined specific heat capacity and demonstrated how it can be used to calculate the heat transferred from or to a system, let’s look at why the specific heat capacity changes between substances.

For example, the specific heat capacity of ethanol is 2.18 joules per gram Kelvin, almost half of water. If we have one gram of water and one gram of ethanol both at 0ºC, it would take 4.18 joules of heat to raise the temperature of water to 1ºC, and only 2.18 joules for ethanol. The liquids reach the same temperature but require different amounts of heat. Why?

Remember, to increase the temperature, we need to increase the average translational kinetic energy of the molecules (make the molecules move faster). But the internal energy of a substance is more than just the translational kinetic energy, it also includes potential energy from intermolecular interactions. When heat is transferred to a system, it is distributed amongst the kinetic and potential energies.

So, if a system has more potential energy, a smaller proportion of the transferred heat is distributed to the kinetic energy, yielding a smaller increase in temperature. To better understand this concept, let’s look at water and ethanol again.

In water, there is a complex network of hydrogen bonds between the molecules. Those interactions are part of the potential energy and need to be overcome, or broken, to increase the average translational kinetic energy. So, when we heat water, some of that energy is used to break up the hydrogen bonding network instead of increasing the kinetic energy, resulting in a large specific heat capacity. Conversely, in ethanol, there are fewer hydrogen bonds per molecule, or less potential energy, and therefore a larger proportion of the heat transferred is used to increase the average kinetic energy, which results in a smaller specific heat capacity.

Review

Okay, let’s wrap up with a review. First, we reviewed the scientific definitions of temperature and heat and related them using specific heat capacity. Using water as an example, we showed how once we know the specific heat capacity, it is quite easy to determine the heat transferred from or to a system. And finally, we considered from a microscopic view why substances have different specific heat capacities.

Thanks for watching and happy studying!

Specific heat capacity refers to the amount of energy or heat required to increase the temperature of 1 gram of a substance by one degree Celsius.

A key characteristic of water is its high specific heat capacity. Water has to absorb 4,184 Joules of heat, or 1 calorie, to increase in temperature of 1 kilogram of water by 1 degree Celsius.

The units often expressed for specific heat capacity are J/(g×C) or J/(kg×C).

Heat capacity, also known as thermal mass, refers to the amount of heat energy require to raise the temperature of an object, and is measure in Joules per Kelvin or Joules per degree Celsius. Specific heat capacity, also known as specific heat, is the heat capacity per unit mass.