Heat-absorbing process

The units are kJ, appropriate for an energy calculation. The drop in temperature of the calorimeter indicates a heat-absorbing process for the calorimeter, so the chemical system absorbs heat from the calorimeter as the compound dissolves. Thus, a positive q is reasonable. Does the magnitude seem reasonable to you ... 

Neither nor Hcomponents can be measured, but their difference, Aft oin, can be readily determined in a constant-pressure calorimeter. Like other enthalpy changes, Aftsoin is positive for endothermic (heat-absorbing) processes and negative for exothermic (heat-generating) processes. 

If an endothermic (heat-absorbing) process occurs, the sample will momentarily become cooler than the reference material the small temperature difference is detected by the pair of thermocouples and a deflection, termed an endotherm, is... 

If the bonds formed are stronger than those broken, the value of AH° is negative and the reaction is defined as exothermic (releasing heat). In contrast, a positive AH° is characteristic of an endothermic (heat-absorbing) process. An example of an exothermic reaction is the combustion of methane, the main component of natural gas, to carbon dioxide and liquid water. 

Thus for isobaric processes a new fimction, the enthalpy H, has been introduced and its change A// is more directly related to the heat that must have been absorbed than is the energy change At/. The same reservations about the meanmg of heat absorbed apply in this process as in the constant-volume process. 

The packaging (qv) requirements for shipping and storage of thermoplastic resins depend on the moisture that can be absorbed by the resin and its effect when the material is heated to processing temperatures. Excess moisture may result in undesirable degradation during melt processing and inferior properties. Condensation polymers such as nylons and polyesters need to be specially predried to very low moisture levels (3,4), ie, less than 0.2% for nylon-6,6 and as low as 0.005% for poly(ethylene terephthalate) which hydrolyzes faster. 

The basic idea of using TCR in a gas turbine is usually to extract more heat from the turbine exhaust gases rather than to reduce substantially the irreversibility of combustion through chemical recuperation of the fuel. One method of TCR involves an overall reaction between the fuel, say methane (CH4), and water vapour, usually produced in a heat recovery steam generator. The heat absorbed in the total process effectively increases... 

To calculate the heat duty it must be remembered that the pressure drop through the choke is instantaneous. That is, no heat is absorbed or lost, but there is a temperature change. This is an adiabatic expansion of the gas w ith no change in enthalpy. Flow through the coils is a constant pressure process, except for the small amount of pressure drop due to friction. Thus, the change in enthalpy of the gas is equal to the heat absorbed. 

Enthalpy changes for biochemical processes can be determined experimentally by measuring the heat absorbed (or given off) by the process in a calorimeter (Figure 3.2). Alternatively, for any process B at equilibrium, the standard-state enthalpy change for the process can be determined from the temperature dependence of the equilibrium constant ... 

Gas or oil fuel is burned in a furnace or boiler (heat exchanger) to heat air or water, or to make steam that carries the heat absorbed from the combustion process to the rooms in the house, as shovra in Figure 2. This IS accomplished by circulating the warm air through ducts directly to the rooms, or by circulating hot water or steam through pipes to baseboard hot water con-... 

For any reversible process, the increase in entropy of any participating system is equal to the heat absorbed by that system divided by the absolute temperature at which the transfer occurred. That is, for a system, i. 

It must be emphasised that the heat q which appears in the definition of entropy (equation 20.137) is always that absorbed (or evolved) when the process is conducted reversibly. If the process is conducted irreversibly and the heat absorbed is q, then q will be less than q, and q/T will be less than AS the entropy change (equation 20.137). It follows that if an irreversible process takes place between the temperatures Tj and 7 , and has the same heat intake q at the higher temperature 7 2 as the corresponding reversible process, the efficiency of the former must be less than that of the latter, i.e. 

In any cyclic process, let a quantity of heat Q be absorbed, and a quantity of work A be done, by the system. Heat emitted, or work done on the system, is to be reckoned with a negative sign. Then the heat absorbed is equivalent to the work done, in the sense explained ... 

Let us now fix our attention on the working substance, i.e. on the material system undergoing the cyclic process. If Qb Q2 are the quantities of heat absorbed by the system from the source and refrigerator respectively ... 

Inequality (7) shows that the heat absorbed from outside is less than corresponds with the increase of entropy of the system (viz., SQ = Td8, where BQ is the heat absorbed in the process when conducted reversibly), hence some entropy is generated inside the system itself. 

Let AU denote the change of intrinsic energy, and let Q be the amount of heat absorbed at the temperature T, in any part of the process. Then, according to the first law ... 

The sum of the amounts of heat absorbed in the four processes, each divided by the temperature at which it is absorbed, is ... 

The isothermal expansion of an ideal gas is an aschistic process.— If a mass of gas expands isothermally, the heat absorbed is equal to the external work done. 

If, therefore, the process is aschistic, i.e., 2A = 2Q, there ought on the whole to be no heat absorbed from or given up to the surroundings ... 

A liquid solution may be separated into its constituents by crystallising out either pure solvent or pure solute, the latter process occurring only with saturated solutions. (At one special temperature, called the cryohydric temperature, both solvent and solute crystallise out side by side in unchanging proportions.) We now consider what happens when a small quantity of solute is separated from or taken up by the saturated solution by reversible processes. Let the saturated solution, with excess of solute, be placed in a cylinder closed below by a semipermeable septum, and the w7hole immersed in pure solvent. The system is in equilibrium if a pressure P, equal to the osmotic pressure of the saturated solution when the free surface of the pure solvent is under atmospheric pressure, is applied to the solution. Dissolution or precipitation of solute can now be brought about by an infinitesimal decrease or increase of the external pressure, and the processes are therefore reversible. If the infinitesimal pressure difference is maintained, and the process conducted so slowly that all changes are isothermal, the heat absorbed when a mol of solute passes into a solution kept always infinitely... 

Suppose that s grams of solute saturate 100 grams of solvent, and let eu c2, e be the specific heats of solute, solvent, and solution, respectively then if Q is the heat absorbed in the process, Kirchhoff s theorem ( 58) shows that... 

The heat absorbed when a mol of the solvent is evaporated at a constant temperature T from a volume of solution so large that no perceptible change of concentration occurs during the process, is called the heat of volatilisation A (x, T). From (7) ... 

We wish to show that no points to the leftbb of 2 on the isotherm 62 are accessible from point 1 via any adiabatic path, reversible or irreversible. Suppose we assume that some adiabatic path does exist between 1 and 2. We represent this path as a dotted curve in Figure 2.11a. We then consider the cycle I —>2 —> 1 — 1. The net heat associated with this cycle would be that arising from the last step 1 — 1, since the other two steps are defined to be adiabatic. We have defined the direction 1 — 1 to correspond to an absorption of heat, which we will call qy. From the first law, the net work vv done in the cycle, is given by w = —q, since AU for the cycle is zero. Thus, for this process, iv is negative (and therefore performed by the system), since qy is positive, having been absorbed from the reservoir. The net effect of this cycle, then, is to completely convert heat absorbed at a high temperature reservoir into work. This is a phenomenon forbidden by the Kelvin-Planck statement of the Second Law. Hence, points to the left of 2 cannot be reached from point 1 by way of any adiabatic path. 

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