Heat and Heat Capacity. Q = mc T

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1 E int = Q + W This is the First Law of Thermodynamics. It says that the internal energy of a system (and its temperature) can be changed by either work being done on (by) the system or heat flowing in (out) of the system. It is a statement of conservation of energy, as applied to the internal energy of the system. Q>0: heat flows into the system due to a difference in temperature with surroundings Q<0: heat flows out of the system due to a difference in temperature with surroundings W>0: done on the system by its surroundings (e.g. it is compressed) W<0: system does work on it surroundings (e.g. it expands) Calorimetry: Changes in materials when no work is done but heat flows between objects. Then for each object: E int = Q [Note: since heat already refers to an energy transfer (i.e. change), we do not write Q.] For small change: de int = dq E int is generally a function of temperature. de int /dt = dq/dt

2 Heat and Heat Capacity If no work is done on the object, de int /dt = dq/dt The material dependent property that describes this is the heat capacity: C(T) dq/dt. [C can be a function of temperature.] Since E int usually increases when the temperature increases, usually C(T) > 0. C is always proportional to amount of material e.g. the mass, so a more intrinsic property is the specific heat: c = C/m = (1/m) dq/dt Q = m c dt If c varies with temperature: Q = mc ave T If c is constant over the temperature range of interest, Q = mc T

3 Water is the champ!

4 Consider two or more objects initially at different temperatures. If they are brought together (i.e. allowed to exchange energy), they will eventually reach the same temperature (thermal equilibrium). [How long this takes depends on how large they are and on how they are connected.] If no work is done by them on each other and no heat flows to or from the outside: Q 1 + Q 2 + Q 3 + Q 4 +. = 0 i.e. whatever heat flows out of some objects (Q j <0) flows into others (Q k >0). m 1 c 1 T 1 + m 2 c 2 T 2 + m 3 c 3 T 3 + m 4 c 4 T 4 + = 0 m 1 c 1 (T f T 1i )+ m 2 c 2 (T f T 2i ) + m 3 c 3 (T f T 3i ) + m 4 c 4 (T f T 4i ) + = 0 where T f is the common final temperature and the T i s are the initial temperatures of each object.

5 Problem: 0.25 kg of water at 95 o C are poured into a 0.1 kg silver cup, initially at 20 o C, that is insulated from the room. What is the final temperature?

6 Problem: 0.25 kg of water at 95 o C are poured into a 0.1 kg silver cup, initially at 20 o C, that is insulated from the room. What is the final temperature? c water = 4186 J/kg o C, c silver = 234 J/kg o C m water c water (T f T water,i ) + m silver c silver (T f -T silver,i ) = 0 or: m silver c silver (T f -T silver,i ) = m water c water (T water,i T f ) [Q into silver = Q out of water] (m water c water + m silver c silver ) T f = m water c water T water,i + m silver c silver T silver,i T f = (m water c water T water,i + m silver c silver T silver,i ) / (m water c water + m silver c silver ) i.e. T f is the weighted averages of thet i s, weighted by their heat capacities (mc s) T f = [(0.25 kg) (4186 J/kg o C) (95 o C) + (0.1 kg) (234 J/kg o C) (20 o C)] / [(0.25 kg) (4186 J/kg o C) + (0.1 kg) (234 J/kg o C)] T f = [(9.942 x x 10 2 ) J] / [( ) (J/ o C)] = 93.4 o C [The water wins because its specific heat is so large.]

7 Problem: 0.2 kg of water at 10 o C is poured into a 0.25 kg glass beaker, initially at 15 o C, that is insulated from the room. Also placed in the beaker is a 0.1 kg piece of aluminum, initially at 150 o C. What is the final temperature? [Note that its not clear if heat will flow in or out of the glass, but you don t need to know that to solve and find T f :]

8 Problem: 0.2 kg of water at 10 o C is poured into a 0.25 kg glass beaker, initially at 15 o C, that is insulated from the room. Also placed in the beaker is a 0.1 kg piece of aluminum, initially at 150 o C. What is the final temperature? c w = 4186 J/kg o C, c a = 900 J/kg o C, c g = 837 J/kg o C m w c w (T f -T w,i ) + m a c a (T f -T a,i ) + m g c g (T f -T g,i ) = 0 T f = [m w c w T w,i + m a c a T a,i + m g c g T g,i ] / [m w c w + m a c a + m g c g ] T f = [(0.2)(4186)(10) + (0.25)(837)(15)+(0.1)(900)(150)] (J) / {[(0.2)(4186) + (0.25)(837)+(0.1)(900)] (J/ o C)} T f = 22.0 o C Note: In these problems, we did not consider changes in temperature of the surrounding air. This is usually OK because the mass of the air is so small. For example, while the specific heat of air is 1000 J/kg o C ¼ c w, the density of air 1.2 kg/m 3 ρ w /1000, so for comparable volumes of air and solid or liquid material, the heat capacity of the air [C air = m air c air ] is negligible.

9 Problem: A 20 kg piece of lead, initially at T = 20 o C falls 100 m into an insulated container holding 2 kg of water initially at 5 o C. If 98% of the kinetic energy of the lead when it hits the water turns into internal energy of the water and lead, what is the final temperature? (The remaining energy may leave the system as sound, for example.) In this problem, some of the temperature change of the water and lead is due to heat flowing between them, but some is due to the input of energy from the falling piece of lead. Conservation of energy: 0.98 K + E int,wat + E int,lead = 0 Now use that for the water and lead, E int = m c T:

10 Problem: A 20 kg piece of lead, initially at T = 20 o C falls 100 m into an insulated container holding 2 kg of water initially at 5 o C. If 98% of the kinetic energy of the lead when it hits the water turns into internal energy of the water and lead, what is the final temperature? (The remaining energy may leave the system as sound, for example.) c wat = 4186 J/kg o C, c lead = 128 J/kg o C Conservation of energy: 0.98 K + E int,wat + E int,lead = 0 K = K f K i = 0 ½ m lead v 2 = - m lead g h -(0.98) m lead g h + m wat c wat (T f T wat,i ) + m lead c lead (T f - T lead,i ) = 0 T f = [0.98 m lead g h + m wat c wat T wat,i + m lead c lead T lead,i ] / ( m wat c wat + m lead c lead ) T f = [(0.98) (20) (9.8) (100) + (2) (4186) (5) + (20) (128) (20)] J / {[(2)(4186) + (20) (128)] J/ o C} T f = ( ) / ( ) o C = o C

11 Room Temperature Properties of Some Materials Material c (J/kg o C) ρ (kg/m 3 ) cρ (10 6 J/m 3 o C) Aluminum Copper Lead Mercury Gold Glass Silicon Marble Ethyl Alcohol Water Note that although values of c vary by more than an order of magnitude, values of cρ are more tightly clustered. That is because in most solids and liquids, the atomic spacing is d 2-3 x m, and cρ = (heat cap/kg) (kg/atom) (atoms/volume) = (heat cap/atom) (atom/volume). Since the number of atoms/ volume is roughly constant, the small variation in cρ means that heat cap/atom does not vary a lot for different materials at room temperature.

12 Material c (J/kg o C) ρ (kg/m 3 ) cρ (10 6 J/m 3 o C) Aluminum Copper Lead Mercury Gold Glass Silicon Marble Ethyl Alcohol Water Although the variation between materials of cρ is much less than that of c alone, water is still the champ. In particular, its value is approximately twice that of glass (or quartz) or marble and other rocks (or earth). Therefore, temperatures change much more slowly over (or near) oceans than inland: coastal areas typically have much more moderate climates than inland areas.

13 Sometimes, the temperature of a material doesn t change even when its internal energy changes. That is, there is no temperature change even when heat flows in (or out), so C = Q/ T = E int / T =. This happens when the physical properties of the material change: such events are called phase changes or phase transitions or phase transformations. Some examples of phase transitions are: a) Melting: solid turning into a liquid. Here the change in E int comes from changes in the bonds between atoms. b) Evaporation (boiling): liquid turns into a gas. Here the change in E int comes from breaking the bonds between atoms. c) Sublimation: solid turns into a gas. Here the change in E int comes from breaking the bonds between atoms. d) Structural changes in solids: Here the change in E int comes from changes in the bonds between atoms. e) Magnetization: temperature at which a ferromagnet stops being magnetic. (For iron, this occurs at 1043 K.) f) Superconductivity: When a metal becomes a perfect conductor (i.e. zero resistance) and perfect diamagnet. For niobium, this occurs at 9.3 K. g) Superfluidity: At 2.2 K, liquid helium beomes a superfluid, i.e. has zero viscocity. (i.e. even less viscous than a gas).

14 These plots shows the phase diagrams of water and CO 2 : i.e. what phase they are in for different temperatures and pressures. [Note that the scales are not linear.] The curves between solid and gas and between liquid and gas give the vapor pressures. The three phases meet at the triple point: at (only) that temperature and pressure, the three phases coexist. The liquid/gas curve ends at the critical point: For T > T cp, the material is called a supercritical fluid. Note that while CO 2 s melting temperature increases with increasing pressure (like most materials), the melting point of ice decreases with increasing pressure. This is related to the fact that liquid water is more dense than solid ice. (For most materials, like CO 2, the solid is more dense than the liquid.)

15 He-II He-I This shows the phase diagram (linear scales) of helium (He 4, the common isotope). Notice the very low critical point = 5.2 K. The normal (non-superfluid) liquid is called He-I and the superfluid phase is called He-II. The transition between them is at the λline, and the vapor pressure becomes extremely small for T < T λ (i.e. P << 1 atm, so can t see on this figure). Also notice that there are two solid phases. The more common structure is hcp (hexagonal close-packed) but there is also a bcc (body-centered cubic) phase in a narrow temperature/pressure region. Also notice that helium does not freeze unless it is pressurized (so there is no sublimation and no solid-liquid-gas triple point). This is because the quantum-mechanical (zero-point) motion of the helium atoms prevents them from ordering in a crystal. Helium is the only material which does not form a solid as P 0.

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