Exam 20: Entropy and the Second Law of Thermodynamics

A perfectly reversible heat pump with a coefficient of performance of 14 supplies energy to a building as heat to maintain its temperature at 27°C.If the pump motor does work at the rate of 1 kW, at what rate does the pump supply energy to the building?

Which of the following is NOT a state variable?

Let S_I denote the change in entropy of a sample for an irreversible process from state A to state B.Let S_R denote the change in entropy of the same sample for a reversible process from state A to state B.Then:

An ideal gas, consisting of n moles, undergoes a reversible isothermal process during which the volume changes from V_i to V_f.The change in entropy of the thermal reservoir in contact with the gas is given by:

For all reversible processes involving a system and its environment:

On a warm day a pool of water transfers energy to the air as heat and freezes.This is a direct violation of:

A slow (quasi-static)process is NOT reversible if:

Consider the following processes: The temperatures of two identical gases are increased from the same initial temperature to the same final temperature.Reversible processes are used.For gas A the process is carried out at constant volume while for gas B it is carried out at constant pressure.The change in entropy:

Let k be the Boltzmann constant.If the thermodynamic state of gas at temperature T changes isothermally and reversibly to a state with three times the number of microstates as initially, the energy input to gas as heat is:

The temperature T_L of the cold reservoirs and the temperatures T_H of the hot reservoirs for four Carnot heat engines are Rank these engines according to their efficiencies, least to greatest

A refrigerator absorbs energy of magnitude as heat from a low temperature reservoir and rejects energy of magnitude as heat to a high temperature reservoir.Work W is done on the working substance.The coefficient of performance is given by:

A Carnot heat engine operates between a hot reservoir at absolute temperature T_H and a cold reservoir at absolute temperature T_L.Its efficiency is:

The change in entropy is zero for:

The difference in entropy $\Delta$ S = S_B - S_A for two states A and B of a system can computed as the integral $\Delta$ dQ/T provided:

A Carnot heat engine runs between a cold reservoir at temperature T_L and a hot reservoir at temperature T_H.You want to increase its efficiency.Of the following, which change results in the greatest increase in efficiency? The value of $\Delta$ T is the same for all changes.

Possible units of entropy are:

For all adiabatic processes:

Which of the following processes leads to a change in entropy of zero for the system undergoing the process?

In a thermally insulated kitchen, an ordinary refrigerator is turned on and its door is left open.The temperature of the room:

The thermodynamic state of gas changes configuration from one with 3.8 *10¹⁸ microstates to one with 7.9 * 10¹⁹ microstates.The Boltzmann constant is 1.38 * 10^-23 J/K.The change in entropy is: