Before moving on to more First Law considerations, heat capacity, and specific heat capacity, we look at the solution of the problem at the end of the previous blog (Basic Physics, Part 12):

(a) The external work done is W = P (V2 - v1) = 1.01 x 10^5 Pa(0.375 - 0.250)m^3

W = 1.25 x 10^4 J, or 12,500 Joules.

(b) delta U = n Cv,m (delta T) = 10(20.2 J/mol K)(T2 - T1), where n = 10 and

For (T2 - T1), we first find T1 = 0 C = 273 + 27 = 300K and we then need to find the higher temperature T2. Since for an isobaric process, V ~ T (P = const.) then:

V2/ V1 = T2/T1 or T2 = (V2/V1) T1 = [300k] x (0.375 m^3)/ (0.250 m^3) = 450 K

Then: (T2 - T1) = (450 K - 300 K) = 150 K, so:

delta U = 10(20.2 J/mol K) (150 K) = 30, 300 J

(c) Heat applied = delta Q = n C p,m (delta T) = 10(28.5 J/mol K) (150 K)

delta Q = 42, 750 J

Let's now go back to reiterate and summarize aspects of the First Law of Thermodynamics, by first noting the types of processes one can obtain under which conditions, given +U = Q - W (another way to express the law, with +U as subject):

(i) Adiabatic process (for Q = 0, i.e. then +U = -W)

(ii) Isobaric process (for P = constant)

(iii) Isovolumetric process (for V = constant)

(iv) Isothermal process (for T = constant)

Other important aspects to note in applying the 1st law:

(a) The conservation of energy statement of the 1st law is independent of path, i.e. (Q - W) is completely determined by the initial and final state, not intermediary states,

Example: say a gas is going from initial state S(i) with P(i), V(i) to final state S(f) with P(f), V(f), then one finds that (Q - W) is the

*same*for all paths connecting S(i) to S(f).

(b) Q is

**positive**(Q > 0) when heat enters the system, and

(c) W is

**positive**(W > 0) when work is done BY the system, and vice versa.

Now, on to heat capacity! This is a generic as opposed to specific quantity defined as the heat that must be transferred to produce a change in temperature, or:

Q = C(T2 - T1). The

*specific heat capacity*, c = C/ m where m is the mass. Then: C = mc and Q = mc (T2 - T1). We also saw already: C' = C/n (= Cp,m, Cv,m) where C' is the

*molar heat capacity*. So, Q= nC' (T2 - T1). The heat capacity has interesting applications apart from prosaic, terrestrial ones.

For example, since space is a near-vacuum, m ~ 0, and c ~ 0, so little or no thermal capacity (C) exists. What this means is that energy from the Sun (via radiation) can be

**transferred through space, without appreciably heating space.**Space is 'cold' not because it absolutely 'lacks heat' but because its density (of particles, hence mass) is too low to have much quantity of heat, or 'thermal capacity'.

What about in the vicinity of Earth? Similar arguments apply. The higher one is above the Earth, the lower the thermal capacity of the medium - so the lower the amount of heat that can be retained, or measured. The lower in altitude one goes, the greater the number of air particles, and the greater retention of heat- especially if water vapor is also included (since water has a large specific heat capacity). What happens is that the radiant energy (mainly from the infrared spectral region) transfers kinetic energy to the molecules of the atmosphere, thereby raising its internal energy: U = 3kT/2. This internal energy, defined along with the thermal capacity of the air(C) is what enables us to feel warmth. Conversely, the relative absence at higher altitudes makes us feel colder.

Specific heat capacity can be measured in simple lab experiments, using the apparatus as set out in the graphic. This image shows: an outer calorimeter (left), inner calorimeter cup (next), and thermometer (far right - inserted in calorimeter cap), and a metal sample for which we may seek to find its specific heat capacity - call it c(x). The practical procedure is then straight forward.

Assuming a mass of water (m_w) say of 100g, and mass of calorimeter (m_cal) which must take the inner cup + outer into account, then we can find what we need using a basic heat conservation equation:

heat lost by hot substance = heat gained by cold substance + heat gained by calorimeter

Then let the unknown metal (mass m_x) be heated to 100 C then deposited in the 200g of water at temperature 27 C in the calorimeter cup (which must be weighed of course). Let the calorimeter be of mass 0.1kg and made of copper for which we know the specific heat capacity c_Cu = 400 J/kg K, then we have:

-m_x c(x) (T - T_x) = m_w c(w)(T - T_w) + m_cal cCu (T - T_w)

This assumes no net heat loss, and also the initial temperature of the calorimeter and the water are the same, e.g. T_w = 20 C. Obviously then, if the unknown metal x is heated to 100 C and dropped into the calorimeter, heat will be gained by both the water and calorimeter, even as heat is lost from the specimen. Now, if c(w) is known to be 4,200 J.kg K we ought to be able to work out the unknown specific heat capacity c(x) if say, the mass m_x is known. Such calorimetric experiments are extremely important since they show several principles at once.

Example problem: For the experimental layout described, let the final temperature attained by the water + calorimeter be 25C. Obtain the unknown specific heat capacity, c(x) if m_x = 0.2 kg:

Then we have: T = 25 C, T_w = 20 C and T_x = 100 C. We also have all other quantities so we can obtain c(x). (Let us also bear in mind here that

*differences in Celsius degrees = differences in Kelvin degrees*). Then we may write:

c(x) = [m_w c(w)(T - T_w) + m_cal cCu (T - T_w)]/[m_x (T - T_x)]

Substituting the measured values of the data:

c(x) = [0.2 kg (4200 J/kg K) (5 K) + 0.1 kg(400 J/kg K)(5 K)/ 0.2 kg( 75 K)

c(x) = [4200 J + 200 J]/ 7.5 kg K = 4400 J/ 15 kg K = 293 J/kg K (which is most likely an alloy of copper and silver, c(Ag) = 234 J/kg K).

Problems:

(1) Let 5 million calories of solar energy be absorbed by 2 cubic meters of hydrogen gas 100 km above the Earth. If the particle density of the gas is 10,000 atoms per cubic meter, estimate the heat capacity of the gas volume. (Atomic mass of a hydrogen atom, 1 u ~ 1.6 x 10^-27 kg).

(2) 10 lbs. of iron and 5 lbs. of aluminum - both at 200 F, are added to 10 lb. of water at 40 F contained in a vessel whose thermal capacity is 0.5 Btu/ deg F. Calculate the final temperature if c(Al) = 0.21 cal/ g C, and c (Fe) = 0.11 cal/g C (Note: specific heat capacities are the same in calories per gram per degree Celsius as in Btu/ F deg).

(3)A calorimeter and its contents have a total thermal (heat) capacity of C = 200 cal/ deg C. A body of mass 210 g and at temperature 80 C is placed in the calorimeter resulting in a temperature increase from 10 C to 20 C. Compute the specific heat capacity of the body.

## No comments:

Post a Comment