Water equivalent of calorimeter Specific latent heat of fusion Open in App Suggest Corrections 0
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? 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? 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. 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. Page 2
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 Click to hide the solution
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 Click to hide the solution 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 per mole = 3277.12J / 0.0083 =394,819J/mole Click to hide the solution
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 b) Calculate the mole and the mass of sugar present.
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% Click to hide the solution joules per degree celsius per gram, J °C-1 g-1 or joules per Kelvin per gram, J K-1 g-1 q = m × Cg × ΔT joules per degree Celsius per mole, J °C-1 mol-1 or joules per Kelvin per mole, J K-1 mol-1 q = n × Cn × ΔT Please do not block ads on this website. Specific Heat CapacityIf you heat some water gently using a heat source like a bunsen burner, the temperature of the water increases. The energy supplied by the bunsen burner causes the water molecules to move faster, increasing their kinetic energy. We can measure the result of this increased kinetic energy as an increase in temperature. The amount of energy absorbed by the water molecules to increase their kinetic energy is referred to as the "heat energy".3 Heat energy of the water particles, q, is proportional to temperature change, ΔT.ΔT = final temperature - initial temperature q ∝ ΔT This means that if you use the same mass of water but double the heat energy (q) then the temperature change (ΔT) will also double. You can could also heat the "cold" water by adding some "hot" water to it. Imagine you have a beaker of water containing 100 g of water at a temperature of 25.0°C. What would happen to the temperature of the water if you added 10 g of boiling water (100°C)? Heat will flow from the hot water to the cold water.4 The kinetic energy of the "hot" water molecules will decrease, and the kinetic energy of the "cold" water molecules will increase, until all the water molecules have the same average kinetic energy.5 Because temperature is a measure of the average kinetic energy of all the water molecules, we find that the temperature of the water will become constant. In this example, a constant temperature6 of 31.8°C will be achieved. The temperature change, ΔT isΔT = final temperature - initial temperature = 31.8 - 25.0 = 6.8°C Now, imagine repeating the experiment, but this time using 20 g of boiling water. What will be the final temperature of the water? Once again heat will flow from the hot water to the cold water, the hot water cools and the cold water heats up until a constant temperature is achieved everywhere in the volume of water. But, this time the temperature will be higher, 37.5°C. The temperature change, ΔT is ΔT = final temperature - initial temperature = 37.5 - 25.0 = 12.5°C Adding a greater mass of hot water to the same mass of cool water raises the temperature more. This tells us that the amount of heat energy that can be transferred from a hot substance to a cold substance is dependent on the mass of the substance used. The heat energy (q) is proportional to the mass of substance used (m) and to the change in temperature (ΔT): q ∝ m × ΔT We could turn this relationship into a mathematical equation by using a constant of proportionality. q = C × m × ΔT Let's see what happens to this constant of proportionality, C, when we change the substance used to heat the water. What would happen to the temperature of 100 g of water initially at 25.0°C if we added 20 g of a different substance instead of water, say, 20 g of copper metal at 100 °C? Heat will flow from the hot copper to the cooler water, the copper will cool down and the water will heat up until a constant temperature is achieved. The final temperature of the water is only 26.5°C, less than the temperature when 20 g of water was added! The temperature change, ΔT is ΔT = final temperature - initial temperature = 26.5 - 25.0 = 1.5°C For equal masses of hot water and hot copper at the same temperature the hot water can transfer more heat energy to the cold water than hot copper metal can.7 That is, the value of the constant of proportionality, C, for water is greater than that for copper. The term that is used to describe this ability (or capacity) to transfer heat energy is "heat capacity". When comparisons are made using the mass in grams of substances, this "heat capacity" is referred to as the specific heat capacity. So, the specific heat capacity of water is greater than the specific heat capacity of copper.Specific heat capacity has been given the symbol Cg (think "g" for grams, that is, mass). Now we can replace the constant of proportionality (C) in our mathematical equation above with specific heat capacity (Cg): q = Cg × m × ΔT We can rearrange this equation by dividing both sides of the equation by m × ΔT:
Now, if I want to compare the specific heat capacities of various substances I would need to keep the mass constant, say 1 gram, and I would use just enough heat energy to produce a temperature change of 1°C (or 1K), That is, the specific heat capacity of a substance is the energy (q) required to raise the temperature of 1 gram of the substance by 1°C (or 1K)! Different substances have different specific heat capacities. The specific heat capacity of some substances is given in the table below:8
From the table above we see that the specific heat capacity of copper is 0.39 J °C-1 g-1 while the specific heat capacity of water is much higher, 4.18 J °C-1 g-1. It requires 0.39 J of energy to change the temperature of 1 gram of copper metal by 1°C (or 1 K). It requires 4.18 J of energy to change the temperature of 1 gram of liquid water by 1°C (or 1 K). Specific heat capacity, Cg, as described above is useful because we can easily measure the mass of many substances. However, when we look at the table of values some of these values seem counter-intuitive. Why should it require 0.13 J of energy to raise the temperature of 1 g of lead 1°C, but almost 7 times as much energy to raise the temperature of 1 g of aluminium by 1°C? And why would carbon have a higher heat capacity than metallic copper or lead? Perhaps comparisons based on mass are not the best option available.....
Do you know this? Join AUS-e-TUTE! Play the game now! Equal masses of different substances contain different numbers of "particles" (atoms, ions, or molecules). The mass of 1 mole of a pure substance is equal to its relative molecular mass expressed in grams: mass of 1 mole = relative molecular mass in grams Recall that specific heat capacity is the energy required to raise the temperature of 1 gram of substance by 1°C (or 1 K). example: Cg for metallic copper, Cu(s), is 0.39 J °C-1 g-1 If we want to find the heat capacity of 1 mole of substance, we need to multiply the specific heat capacity, Cg, by the relative molecular mass (Mr) or molar mass (M) of the substance: heat capacity of 1 mole = Mr × C(g) or heat capacity of 1 mole = M × C(g) The quantity "M × Cg" is referred to as the molar heat capacity and is given the symbol Cn (n is the symbol used for moles). The molar heat capacity of a substance is the energy required to raise the temperature of 1 mole of substance by 1°C (or 1K). For example, the specific heat capacity of copper metal: Cg = 0.39 J°C-1 g-1 You could perform this calculation yourself for each of the substances listed in the table of specific heat capacities above.
This table allows us to compare the heat capacities of the same number of particles, that is, 1 mole, of different substances. We find that the molar heat capacities of the metals is quite similar while the molar heat capacity of carbon is much lower. It requires about 25 J of energy to raise the temperature of 1 mole of metal by 1°C (or 1 K), but it only needs about 9 J of energy to raise the temperature of 1 mole of carbon by 1°C (or 1 K). We could write a new equation for calculating the amount of heat required (q) to raise the temperature (ΔT) of an amount of substance in moles (n): q = Cn × n × ΔT
Do you understand this? Join AUS-e-TUTE! Take the test now! Question 1: Calculate the quantity of heat in joules needed to increase the temperature of 250 g of water from 20°C to 56°C. Solution: (Based on the StoPGoPS approach to problem solving.)
ΔT = Tf - Ti (b) Substitute the values for m, Cg and ΔT into the equation and solve for q q = m × Cg × ΔT Question 2: Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15 g of copper from 25°C to 60°C. Solution: (Based on the StoPGoPS approach to problem solving.)
Question 3: 216 J of energy is required to raise the temperature of aluminium from 15°C to 35°C. Calculate the mass in grams of aluminium. (Specific Heat Capacity of aluminium is 0.90 J°C-1g-1). Solution: (Based on the StoPGoPS approach to problem solving.)
Question 4: The initial temperature of 150 g of ethanol was 22.0°C. What will be the final temperature in degrees celsius of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (Specific heat capacity of ethanol is 2.44 J°C-1g-1) Solution: (Based on the StoPGoPS approach to problem solving.)
(b) Calculate the final temperature of ethanol: Tf = ΔT + Ti
Can you apply this? Join AUS-e-TUTE! Take the exam now! 1. Since the graduations in the scale of both the celsius and Kelvin temperature scales are the same, and because we are not concerned here with either the initial or final temperature but only in the difference between the two, it can been seen that a difference of 1°C is the same as a difference of 1 K. 2. In 1960 the General Conference of Weights and Measures agreed upon a unified version of the metric system. The units in this system are known as SI units (Systèm International d'Unités). Seven base units constitute the foundation of the SI system:
The unit of force is the newtown (N), it is a derived unit, 1 N = 1 kg m s-2 The unit for energy is also a derived unit, the joule (J), 1 J = 1 N m = 1 kg m2 s-2 Electrical measurements are capable of greater precision than calorimetric measurements as described in this discussion, so the joule can also be defined as a volt coulomb. 3. Heat, or heat energy, is the energy directly transferred from one object to another. Heat is energy in transit, a substance like water at a constant temperature does not have a "heat content", but it does have an "energy content". The energy content of a substance is made up of the kinetic energy (movement) of its particles and potential energy such as the stored chemical potential energy in its chemical bonds. Temperature is a measure of the average kinetic energy of the particles. 4. Heat always flows from "hot" to "cold". 5. The particles will not have exactly the same kinetic energy. There is a distribution of kinetic energies for the particles, so we refer to the "average" kinetic energy of the particles in the system. 6. This is known as thermal equilibrium. 7. It is more accurate to say that the heat capacity is the ability of a substance to transfer heat to another substance since heat is energy in transit. That is, heat capacity is the capacity or ability of a substance to transfer heat to another substance. But, since the word capacity is usually thought of as "containment", eg, a 250 mL volumetric flask has a capacity of 250 mL, we often think of the heat capacity of a substance being its ability to contain heat energy. We cannot really refer to "heat" as being stored, that is, heat can be absorbed by the molecules to increase their kinetic energy but it is not "stored" because it has done work to speed up the particles. Heat energy could be "stored" as potential energy in chemical bonds if a chemical reaction takes place, but this is not the case in these examples. 8. Values of specific heat capacities refer to conditions of constant atmospheric pressure. |