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Measuring The Quantity Of Heat
Heat gains or losses result in changes in temperature, changes in state or the performance of work. When gained or lost by an object, there will be corresponding energy changes within that object. A change in temperature is associated with changes in the average kinetic energy of the particles within the object.
A change in state is associated with changes in the internal potential energy possessed by the object. And when work is done, there is an overall transfer of energy to the object upon which the work is done. In this Lesson , we will investigate the question How does one measure the quantity of heat gained or released by an object?
Specific and Latent Heat Capacity
Suppose that several objects composed of different materials are heated in the same manner. Will the objects warm up at equal rates? The answer: most likely not. Different materials would warm up at different rates because each material has its own specific heat capacity.
The specific heat capacity refers to the amount of heat required to cause a unit of mass (say a gram or a kilogram) to change its temperature by 1°C. Specific heat capacities of various materials are often given in Joules/kilogram/Kelvin (J/kg/K). More commonly used units are J/g/°C.
The specific heat capacity of solid aluminum (0.904 J/g/°C) is different than the specific heat capacity of solid iron (0.449 J/g/°C). This means that it would require more heat to increase the temperature of a given mass of aluminum by 1°C compared to the amount of heat required to increase the temperature of the same mass of iron by 1°C. In fact, it would take about twice as much heat to increase the temperature of a sample of a given amount of aluminum compared to the same temperature change of the same amount of iron. This is because the specific heat capacity of aluminum is nearly twice the value of iron.
Latent heat (also known as latent energy, or as Heat of Transformation) is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process — usually a first-order phase transition. Latent heat can be understood as energy in hidden form which is supplied or extracted to change the state of a substance without changing its temperature. Examples are latent heat of fusion and latent heat of vaporization involved in phase changes, i.e. a substance condensing or vaporizing at a specified temperature and pressure. In contrast to latent heat, sensible heat is energy transferred as heat, with a resultant temperature change in a body.
The energy required for a particular change in temperature is given by the specific heat capacity.The energy required for a particular change in state is given by the specific latent heat
Heat capacities are listed on a per gram or per kilogram basis. Occasionally, the value is listed on a per mole basis, in which case it is called the molar heat capacity. The fact that they are listed on a per amount basis is an indication that the quantity of heat required to raise the temperature of a substance depends on how much substance there is.
Water boils at 100°C at sea level and at slightly lowered temperatures at higher elevations. To bring a pot of water to a boil, its temperature must first be raised to 100°C. This temperature change is achieved by the absorption of heat from the stove burner. One quickly notices that it takes considerably more time to bring a full pot of water to a boil than to bring a half-full of water to a boil. This is because the full pot of water must absorb more heat to result in the same temperature change. In fact, it requires twice as much heat to cause the same temperature change in twice the mass of water.
When that energy is transferred to other objects of different temperatures, we refer to transferred energy as heat or thermal energy.
Relating the Quantity of Heat to the Temperature Change
Specific heat capacities provide a means of mathematically relating the amount of thermal energy gained (or lost) by a sample of any substance to the sample’s mass and its resulting temperature change. The relationship between these four quantities is often expressed by the following equation.
Q = m•C•ΔT
where Q is the quantity of heat transferred to or from the object, m is the mass of the object, C is the specific heat capacity of the material the object is composed of, and ΔT is the resulting temperature change of the object.
As in all situations in science, a delta (∆) value for any quantity is calculated by subtracting the initial value of the quantity from the final value of the quantity. In this case, ΔT is equal to Tfinal – Tinitial.
When using the above equation, the Q value can turn out to be either positive or negative. As always, a positive and a negative result from a calculation has physical significance. A positive Q value indicates that the object gained thermal energy from its surroundings; this would correspond to an increase in temperature and a positive ΔT value.
A negative Q value indicates that the object released thermal energy to its surroundings; this would correspond to a decrease in temperature and a negative ΔT value.
Knowing any three of these four quantities allows an individual to calculate the fourth quantity. A common task in many physics classes involves solving problems associated with the relationships between these four quantities. As examples, consider the two problems below.
Example Problem 1
What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C.
Like any problem in physics, the solution begins by identifying known quantities and relating them to the symbols used in the relevant equation. In this problem, we know the following:
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
We wish to determine the value of Q – the quantity of heat. To do so, we would use the equation Q = m•C•ΔT. The m and the C are known; the ΔT can be determined from the initial and final temperature.
T = Tfinal – Tinitial = 85°C – 15°C = 70.°C
With three of the four quantities of the relevant equation known, we can substitute and solve for Q.
Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C)
Q = 131670 J
Q = 1.3×105 J = 130 kJ (rounded to two significant digits)
Example Problem 2
A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.
Compared to the previous problem, this is a much more difficult problem. In fact, this problem is like two problems in one. At the center of the problem-solving strategy is the recognition that the quantity of heat lost by the water (Qwater) equals the quantity of heat gained by the metal (Qmetal).
Since the m, C and ΔT values of the water are known, the Qwater can be calculated. This Qwater value equals the Qmetal value. Once the Qmetal value is known, it can be used with the m and ΔT value of the metal to calculate the Qmetal. Use of this strategy leads to the following solution:
Part 1: Determine the Heat Lost by the Water
- Given:
- m = 50.0 g
C = 4.18 J/g/°C
Tinitial = 88.6°C
Tfinal = 87.1°C
ΔT = -1.5°C (Tfinal – Tinitial) - Solve for Qwater:
- Qwater = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)
Qwater = -313.5 J (unrounded)
(The – sign indicates that heat is lost by the water) - Part 2: Determine the value of Cmetal
- Given:
- Qmetal = 313.5 J (use a + sign since the metal is gaining heat)
m = 12.9 g
Tinitial = 26.5°C
Tfinal = 87.1°C
ΔT = (Tfinal – Tinitial ) - Solve for Cmetal:
- Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain Cmetal = Qmetal / (mmetal•ΔTmetal)
- Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9 g)•(60.6°C)]
Cmetal = 0.40 X103 J/g/°C
Cmetal = 0.40 J/g/°C (rounded to two significant digits)