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Elementary Thermodynamics - Specific Heat and Enthalpy

Earlier, we saw how units of heat measurement were defined in terms of the amount of energy gain or loss required to alter the temperature of a water sample by one degree. In the case of the calorie, it was the amount of heat gain/loss required to heat/cool one gram of water one degree Celsius. In the case of the BTU, it was the amount of heat gain/loss required to heat/cool one pound of water one degree Fahrenheit.

As one might expect, one heat unit might be similarly defined as the amount of heat gain or loss to alter the temperature one-half of a degree for twice as much water, or two degrees for half as much water. We could express this as a proportionality:


Where,

  Q = Heat gain or loss

  m = Mass of sample

  ΔT = Temperature change (rise or fall) over time

 

The next logical question to ask is, “How does the relationship between heat and temperature change work for substances other than water?” Does it take the same amount of heat to change one gram of iron by one degree Celsius as it does water? The answer to this question is a resounding no! Different substances require vastly different amounts of heat gain/loss to alter their temperature by the same amount, even when the masses of those substances are identical.

We have a term for this ability to absorb or release heat, called heat capacity or specific heat, symbolized by the variable c. Thus, our heat/mass/temperature change relationship may be described as a true formula instead of a mere proportionality:

Where,

  Q = Heat gain or loss (metric calories or British BTU)

  m = Mass of sample (metric grams or British pounds)

  c = Specific heat of substance

  ΔT = Temperature change (metric degrees Celsius or British degrees Fahrenheit)

 

Pure water, being the standard by which all other substances are measured, has a specific heat value of 1. The smaller the value for c, the less heat gain or loss is required to alter the substance’s temperature by a set amount. That substance (with a low value of c) has a low “heat capacity” because each degree of temperature rise or fall represents a relatively small amount of energy gained or lost. Substances with low c values are easy to heat and cool, while substances having high c values require much heat in order to alter their temperatures, assuming equal masses.

A table of specific heat values (at room temperature, 25 degrees Celsius1) for common substances appears here:


  Substance 
  Specific heat value (c) cal/goC or BTU/lboF  
  Aluminum (solid) 
  0.215 
  Iron (solid) 
  0.108 
  Copper (solid) 
  0.092 
  Lead (solid) 
  0.031 
   Water (liquid) 
  1.0 
  Methanol (liquid) 
  0.609 
  Ethanol (liquid) 
  0.587 
  Acetone (liquid) 
  0.521 
  Hydrogen (gas) 
  3.41 
  Helium (gas) 
  1.24 
  Nitrogen (gas) 
  0.249 
  Oxygen (gas) 
  0.219 

 If a liquid or a gas is chosen for use as a coolant (a substance to efficiently convect heat away from an object), greater values of c are better. Water is one of the best liquid coolants with its relatively high c value of one: it has more capacity to absorb heat than other liquids, for the same rise in temperature. The ideal coolant would have an infinite c value, being able to absorb an infinite amount of heat without itself rising in temperature at all.

As you can see from the table, the light gases (hydrogen and helium) have extraordinarily high c values. Consequently, they function as excellent gas coolants. This is why large electric power generators often use hydrogen gas as a coolant: that gas has an amazing ability to absorb heat from the wire windings of a generator without suffering a large rise in temperature. Helium, although not as good a coolant as hydrogen, has the distinct advantage of being chemically inert (non-reactive), in stark contrast to hydrogen’s extreme flammability. Some nuclear reactors use helium as a coolant rather than a liquid such as water or liquefied sodium metal.

Numerical examples are helpful in better understanding specific heat. Consider a case where a copper pot filled with water receives heat from a small gas burner operating at an output of 5,000 BTU per hour (350 calories per second):


A reasonable question to ask would be, “How much will the temperature of this water-filled pot rise after 40 seconds of heating?” With the burner’s heat output of 350 calories per second and a heating time of 40 seconds, we may assume2 the amount of heat absorbed by the water-filled pot will be the simple product of heat rate times time:


This amount of heat not only goes into raising the temperature of the water, but it also raises the temperature of the copper pot. Each substance (water, copper) has its own specific heat and mass values (c and m), but they will share the same temperature rise (ΔT), so we must sum their heats as follows:


Solving this equation for temperature rise, we get:


So, if the water and pot began at a temperature of 20 degrees Celsius, they will be at a temperature of 23.68 degrees Celsius after 40 seconds of heating over this small burner. Another example involves the mixing of two substances at different temperatures. Suppose a heated mass of iron drops into a cool container3 of water. Obviously, the iron will lose heat energy to the water, causing the iron to decrease in temperature while the water rises in temperature. Suppose the iron’s mass is 100 grams, and its original temperature is 65 degrees Celsius. Suppose the water’s mass is 500 grams, and its original temperature is 20 degrees Celsius:


What will the equilibrium temperature be after the iron falls into the water and both their temperatures equalize? We may solve this by setting two heat equations equal to each other4: the heat lost by the iron and the heat gained by the water, with the final equilibrium temperature being T:

 

Note how the ΔT term is carefully set up for each side of the equation. In order to make the iron’s heat loss a positive value and the water’s heat gain a positive value, we must ensure the quantity within each set of parentheses is positive. For the iron, this means ΔT will be 65 degrees minus the final temperature. For the water, this means ΔT will be the final temperature minus its starting temperature of 20 degrees.

 

Thus, the iron’s temperature falls from 65 degrees Celsius to 20.95 degrees Celsius, while the water’s temperature rises from 20 degrees Celsius to 20.95 degrees Celsius. The water’s tremendous specific heat value compared to the iron (nearly 10 times as much!), as well as its superior mass (5 times as much) results in a much larger temperature change for the iron than for the water.

 

An analogy to help grasp the concept of specific heat is to imagine heat as a fluid5 that may be “poured” into vessels of different size, those vessels being objects or substances to be heated. The amount of liquid held by any vessel represents the total amount of thermal energy, while the height of the liquid inside any vessel represents its temperature:

The factor determining the relationship between liquid volume (heat) and liquid height (temperature) is of course the cross-sectional area of the vessel. The wider the vessel, the more heat will be required to “fill” it up to any given temperature. In this analogy, the area of the vessel is analogous to the term mc: the product of mass and specific heat. Objects with larger mass require more heat to raise their temperature to any specific point, specific heats being equal. Likewise, objects with large specific heat values require more heat to raise their temperature to any specific point, masses being equal.

In the first numerical calculation example where we determined the temperature of a pot of water after 40 seconds of heating, the analogous model would be to determine the height of liquid in a vessel after pouring liquid into it for 40 seconds at a fixed rate. A model for the second numerical example would be to calculate the equilibrium height (of liquid) after connecting two vessels together at their bottoms with a tube. Although the liquid heights of those vessels may be different at first, the levels will equalize after time by way of liquid passing through the tube from the higher-level vessel to the lower-level vessel.

Many industrial processes use fluids to convectively transfer heat from one object (or fluid) to another. In such applications, it is important to know how much heat will be carried by a specific quantity of that fluid over a specified temperature drop. One common way to express this heat quantity is called enthalpy. Enthalpy is the amount of heat lost by a unit mass (one gram metric, or one pound British) of the fluid as it cools from a given temperature all the way down to the freezing point of water (0 degrees Celsius, or 32 degrees Fahrenheit). A sample of water at a temperature of 125 degrees Fahrenheit, for example, has an enthalpy of 93 BTU per pound (or 93 calories per gram):

 

Even if the process in question does not cool the heat transfer fluid down to water’s freezing point, enthalpy is a useful figure for estimating the thermal energy “content” of hot fluids (per unit mass). Enthalpy is especially useful when dealing with heat transfer fluids as they change phase from vapor to liquid, as the following subsection will discuss.

 

1An important detail to note is that specific heat does not remain constant over wide temperature changes. This complicates calculations of heat required to change the temperature of a sample: instead of simply multiplying the temperature change by mass and specific heat (Q = mcΔT or Q = mc[T2 T1]), we must integrate specific heat over the range of temperature ( ), summing up infinitesimal products of specific heat and temperature change (c dT ) over the range starting from temperature T1 through temperature T2 then multiplying by the mass to calculate total heat required.

2In reality, the amount of heat actually absorbed by the pot will be less than this, because there will be heat losses from the warm pot to the surrounding (cooler) air. However, for the sake of simplicity, we will assume all the burner’s heat output goes into the pot and the water it holds.

3We will assume for the sake of this example that the container holding the water is of negligible mass, such as a Styrofoam cup. This way, we do not have to include the container’s mass or its specific heat into the calculation.

4An alternative way to set up the problem would be to calculate ΔT for each term as Tfinal Tstart, making the iron’s heat loss a negative quantity and the water’s heat gain a positive quantity, in which case we would have to set up the equation as a zero-sum balance, with zero equal to Qiron + Qwater. I find this approach less intuitive than simply saying the iron’s heat loss will be equal to the water’s heat gain, and setting up the equation as two positive values equal to each other.

5This is not far from the reality of eighteenth-century science, where heat was thought to be an invisible fluid called caloric.

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