| •The amount of energy required to raise the temperature of one gram of a substance by 1°C is called the specific heat capacity of that substance.The higher the specific heat capacity, the more effectively a material will store heat energy.•Heat change gained or lost by a substance during a chemical reaction can be calculated by: E = m x c x ΔT where E is Energy (J); m is Mass (g); c is specific heat capacity (Jg-1oC-1); T is temperature (oC), Δ means change in.Example •Calculate the energy required to heat 120mL of water for a cup of coffee to boiling point if the initial water temperatuer is 20.0°C.Since the density of water is 1g mL, the mass of 120mL is 120g. E = m x c x ΔTEnergy required to raise temperature of 120g of water by 1 degree = SHC x mass4.184 x 120 = 502.0JSince temperature rises by 80 degrees, the total energy required = J x ΔT= 502.0 x 80 = 40160J = 40.2kJ •Enthalpy changes are measured directly using an instrument call a calorimeter.•The picture on the right shows the components of a bomb calorimeter used for reactions that involved gaseous reactants or products.•The reaction vessel is designed to withstand high pressures created during the reaction.•The second picture, is a calorimeter used for reactions in aqueous solutions.•Both calorimeters are insulated to reduce loss or gain of energy to or from the outside environment. •When a reaction takes place in a calorimeter, the heat change causes a rise or fall in the temperature of the contents of the calorimeter.•Before the calorimeter can be of use, it must first be determined how much energy is required to change the temperature within the calorimeter by 1°C.•This is known as the calibration factor of the calorimeter.•The calorimeter is calibrated by using an electric heater to release a known quantity of thermal energy and measuring the resultant rise in temperature.•The thermal energy released when an electric current passes through the heater can be calculated from the formula:Energy = voltage (volts) x current(amps) x time (seconds) OR; E = VIt Alternatively Energy can be calculated using:E = n xΔH for a known ΔH value THE HEAT OF COMBUSTION •The heat of combustion of a substance is defined as the energy released when a specified amount (eg. 1 mol, 1g, 1 L) of the substance burns completely in oxygen.•Heats of combustion are measured using a calorimeter.•ΔHc are given for substances in the data booklet (See below).
Aus-e-tute specific heat capacity Aus-e-tute heat of enthalpy combustion Aus-e-tute enthalpy of reaction Aus-e-tute Enthalpy of solutions Aus-e-tute Enthalpy of neutralisation | Diagram By Akshat Goel - Own work, CC BY-SA 3.0, //commons.wikimedia.org/w/index.php?curid=18702902
Bomb calorimeter diagram By Lisdavid89 - Own work, CC BY-SA 3.0, //commons.wikimedia.org/w/index.php?curid=22537546 •The calibration factor is determined using: cf = Energy added where cf = calibration factor ΔT calibrationThis can be used to find the energy of reaction: Energy reaction = cf x ΔT reaction FINDING ΔH FROM CALORIMETRY ΔH can be determined using the following relationship: ΔH = E reaction EXAMPLE •A bomb calorimeter was calibrated by passing 1.5A through the electric heater for 50.0s at a potential difference of 5.43V. The temperature of the water in the calorimeter rose by 0.412°C.ΔH for the equation: CH4(g) + 2O2(g) à CO2(g) + 2H2O(l)was determined by burning 9.21x10-4mol of methane gas in the calorimeter. The temperature rose from 20.40°C to 21.23°C.Step 1: Determine the calibration factor of the calorimeter using E = VIt E= 5.43V x 1.50A x 50.)s = 406.94JSince this energy made the temperature rise by 0.387°C, the energy required to raise the temperature by 1°C = 406.94/0.412 = 987.71J°C-1.Step 2: Calculate the energy change during the reaction. ΔT = 21.23 – 20.40 = 0.83°CEnergy change in calorimeter = calibration factor x ΔT= 987.71x0.83 = 819.8JStep 3: Calculate ΔH for the equation. Note- mole ratio of 1, n1 = 1ΔH = Energy change / mol = 819.8 / (9.21x10-4) = 890119JΔH = -890kJmol-1 (-ve because it indicates that energy has been released, explaining the increase in temperature. |
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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