What is the quantity of heat required to raise the temperature of 1 gram of water through 1°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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REET 2022 Written Exam Result Out on 29th September 2022! The final answer key was also out with the result. The exam was conducted on 23rd and 24th July 2022. The candidates must go through the REET Result 2022 to get the direct link and detailed information on how to check the result. The candidates who will be finally selected for 3rd Grade Teachers are expected to receive Rs. 23,700 as salary. Then, the candidates will have to serve the probation period which will last for 2 years. Also, note during probation, the teachers will receive only the basic salary.