previous index next Show Measuring Heat Flow: a Unit of HeatThe next obvious question is, can we get more quantitative about this “flow of heat” that takes place between bodies as they move towards thermal equilibrium? For example, suppose I reproduce one of Fahrenheit’s experiments, by taking 100 ccs of water at 100°F, and 100 ccs at 150°F, and mix them together in an insulated jug so little heat escapes. What is the final temperature of the mix? Of course, it’s close to 125°F—not surprising, but it does tell us something! It tells us that the amount of heat required to raise the temperature of 100 cc of water from 100°F to 125°F is exactly the same as the amount needed to raise it from 125°F to 150°F. A series of such experiments (done by Fahrenheit, Black and others) established that it always took the same amount of heat to raise the temperature of 1 cc of water by one degree. This makes it possible to define a unit of heat. Perhaps unfairly to Fahrenheit, 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 degree Celsius. (Warning: the food Calorie, with a capital C, is 1000 calories, it's the heat to raise one kilogram of water one degree C. It's sometimes written kilocalorie, abbreviated kCal.) (Celsius also lived in the early 1700’s. His scale has the freezing point of water as 0°C, the boiling point as 100°C. Fahrenheit’s scale is no longer used in science, but lives on in engineering in the US, and in the British Thermal Unit, which is the heat required to raise the temperature of one pound of water by 1°F.) Specific Heats and CalorimetryFirst, let’s define specific heat: The specific heat of a substance is the heat required in calories to raise the temperature of 1 gram by 1 degree Celsius. As Fahrenheit continues his measurements of heat flow, it quickly became evident that for different materials, the amount of heat needed to raise the temperature of one gram by one degree could be quite different. For example, it had been widely thought before the measurements were made, that one cc of Mercury, being a lot heavier than one cc of water, would take more heat to raise its temperature by one degree. This proved not to be the case—Fahrenheit himself made the measurement. In an insulating container, called a “calorimeter” he added 100 ccs of water at 100°F to 100 ccs of mercury at 150°F, and stirred so they quickly reached thermal equilibrium. Question: what do you think the final temperature was? Approximately? Answer: The final temperature was, surprisingly, about 120°F. 100 cc of water evidently “contained more heat” than 100 cc of mercury, despite the large difference in weight! This technique, called calorimetry, was widely used to find the specific heats of many different substances, and at first no clear pattern emerged. It was puzzling that the specific heat of mercury was so low compared with water! As more experiments on different substances were done, it gradually became evident that heavier substances, paradoxically, had lower specific heats. A Surprising Connection With Atomic TheoryMeanwhile, this quantitative approach to scientific observation had spread to chemistry. Towards the end of the 1700’s, Lavoisier weighed chemicals involved in reactions before and after the reaction. This involved weighing the gases involved, so had to be carried out in closed containers, so that, for example, the weight of oxygen used and the carbon dioxide, etc., produced would be accounted for in studying combustion. The big discovery was that mass was neither created nor destroyed. This had not been realized before because no one had weighed the gases involved. It made the atomic theory suddenly more plausible, with the idea that maybe chemical reactions were just rearrangements of atoms into different combinations. Lavoisier also clarified the concept of an element, an idea that was taken up in about 1800 by John Dalton, who argues that a given compound consisted of identical molecules, made up of elementary atoms in the same proportion, such as H2O (although that was thought initially to be HO). This explained why, when substances reacted chemically, such as the burning of hydrogen to form water, it took exactly eight grams of oxygen for each gram of hydrogen. (Well, you could also produce H2O2 under the right conditions, with exactly sixteen grams of oxygen to one of hydrogen, but the simple ratios of amounts of oxygen needed for the two reactions were simply explained by different molecular structures, and made the atomic hypothesis even more plausible.) Much effort was expended carefully weighing the constituents in many chemical reactions, and constructing diagrams of the molecules. The important result of all this work was that it became possible to list the relative weights of the atoms involved. For example, the data on H2O and H2O2 led to the conclusion that an oxygen atom weighed sixteen times the weight of a hydrogen atom. It must be emphasized, though, that these results gave no clue as to the actual weights of atoms! All that was known was that atoms were too small to see in the best microscopes. Nevertheless, knowing the relative weights of some atoms in 1820 led to an important discovery. Two professors in France, Dulong and Petit, found that for a whole series of elements the product of atomic weight and specific heat was the same!
The significance of this, as they pointed out, was that the “specific heat”, or heat capacity, of each atom was the same—a piece of lead and a piece of zinc having the same number of atoms would have the same heat capacity. So heavier atoms absorbed no more heat than lighter atoms for a given rise in temperature. This partially explained why mercury had such a surprisingly low heat capacity. Of course, having no idea how big the atoms might be, they could go no further. And, indeed, many of their colleagues didn’t believe in atoms anyway, so it was hard to convince them of the significance of this discovery. Latent HeatOne of Black’s experiments was to set a pan of water on a steady fire and observe the temperature as a function of time. He found it steadily increased, reflecting the supply of heat from the fire, until the water began to boil, whereupon the temperature stayed the same for a long time. The steam coming off was at the same (boiling) temperature as the water. So what was happening to the heat being supplied? Black correctly concluded that heat needed to be supplied to change water from its liquid state to its gaseous state, that is, to steam. In fact, a lot of heat had to be supplied: 540 calories per gram, as opposed to the mere 100 calories per gram needed to bring it from the freezing temperature to boiling. He also discovered that it took 80 calories per gram to melt ice into water, with no rise in temperature. This heat is released when the water freezes back to ice, so it is somehow “hidden” in the water. He called it latent heat, meaning hidden heat. Books I used in preparing this lecture: A Source Book in Greek Science, M. R. Cohen and I. E. Drabkin, Harvard university Press, 1966. A History of the Thermometer and its Uses in Meteorology, W. E. Knowles Middleton, Johns Hopkins Press, 1966. A Source Book in Physics, W. F. Magie, McGraw-Hill, New York, 1935. previous index next
This worked example problem demonstrates how to calculate the energy required to raise the temperature of a sample that includes changes in phase. This problem finds the energy required to turn cold ice into hot steam. What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?Useful information:heat of fusion of water = 334 J/gheat of vaporization of water = 2257 J/gspecific heat of ice = 2.09 J/g·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°C The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up. Step 1: Find the heat required to raise the temperature of ice from -10 °C to 0 °C. Use the formula: q = mcΔT where
In this problem:
Plug in the values and solve for q: q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]q = (25 g)x(2.09 J/g·°C)x(10 °C) q = 522.5 J
Find the heat required to convert 0 °C ice to 0 °C water.
q = m·ΔHf where
For this problem:
Plugging in the values gives the value for q:
q = (25 g)x(334 J/g) The heat required to convert 0 °C ice to 0 °C water = 8350 J
Find the heat required to raise the temperature of 0 °C water to 100 °C water.q = mcΔTq = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]q = (25 g)x(4.18 J/g·°C)x(100 °C)q = 10450 JThe heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J Step 4:
Find the heat required to convert 100 °C water to 100 °C steam. q = heat energy m = massΔHv = heat of vaporization q = (25 g)x(2257 J/g)q = 56425 JThe heat required to convert 100 °C water to 100 °C steam = 56425 Step 5: Find the heat required to convert 100 °C steam to 150 °C steamq = mcΔTq = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]q = (25 g)x(2.09 J/g·°C)x(50 °C)q = 2612.5 J The heat required to convert 100 °C steam to 150 °C steam = 2612.5 Step 6: Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.
Answer: The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.
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How many calories would it take to change the state of 1 gram of water from 100 degrees C to steam at 100 degrees C?
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