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Chapter 15: 17Q (page 412)

Question: Which do you think has the greater entropy, 1 kg of solid iron or 1 kg of liquid iron? Why?

Short Answer

Expert verified

The entropy of the one kg liquid iron is greater than the entropy of the one kg solid iron.

Step by step solution

01

Meaning of entropy

The term entropy may be defined as the characteristic of a system that estimates the system's degree of disorder. It is considered as the state function.

02

Entropy of a liquid and solid substance 

The order of composition of molecules in liquid iron is weaker than the order of solid iron. Hence, the molecules have higher kinetic energy in liquid iron than solid iron and possess more thermal energy. Thus, the entropy of liquid iron is greater than solid iron.

To melt the solid iron, we must provide heat energy. The provision of heat enhances the kinetic energy of molecules and hence improves their entropy.

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Most popular questions from this chapter

Question:(II) A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%.

Question: (II) (a) What is the coefficient of performance of an ideal heat pump that extracts heat from 6掳C air outside and deposits heat inside a house at 24掳C? (b) If this heat pump operates on 1200 W of electrical power, what is the maximum heat it can deliver into the house each hour? See Problem 35.

Question:(II) A heat engine uses a heat source at 580掳C and has an ideal (Carnot) efficiency of 22%. To increase the ideal efficiency to 42%, what must be the temperature of the heat source?

A 2.8-kg piece of aluminum at 28.5掳C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20.0掳C). Estimate the net change in entropy of the system.

Question: (III) The PV diagram in Fig. 15鈥�23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.
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