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An aluminum rod conducts 8.40 cal/s from a heat source maintained at 225°C to a large body of water at 22°C. Calculate the rate at which entropy increases in this process.

Short Answer

Expert verified

The rate of increase of entropy is \(4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\).

Step by step solution

01

Concepts

The extracted heat from the heat source is coming at a constant rate.

The change in entropy is\(\Delta S = \frac{Q}{T}\)

02

Given data

The rate of heat transfer is \(Q = 8.40\;{\rm{cal/s}} = \left( {8.40 \times 4.186} \right)\;{\rm{J/s}}\).

The sourece temparature is \({T_1} = {225 \circ }{\rm{C}} = 498\;{\rm{K}}\).

The lower temperature is \({T_2} = {22 \circ }{\rm{C}} = 295\;{\rm{K}}\).

03

Calculation

The same amount of heat comes out from the heat source and is absorbed by the lower-temperature body.

Now the rate of change in entropy during the heat coming from the source is \(\Delta {S_1} = - \frac{Q}{{{T_1}}}\).

The rate of change in entropy during the absorption of heat is \(\Delta {S_2} = \frac{Q}{{{T_2}}}\).

Therefore the rate of change of net entropy is,

\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = - \frac{Q}{{{T_1}}} + \frac{Q}{{{T_2}}}\\ = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}\Delta S = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\\ = \left[ {\left( {8.40 \times 4.186} \right)\;{\rm{J/s}}} \right] \times \left( {\frac{1}{{295\;{\rm{K}}}} - \frac{1}{{498\;{\rm{K}}}}} \right)\\ = 4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\end{array}\)

Hence, the rate of increase of entropy is \(4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\).

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Most popular questions from this chapter

Question: (II) A heat pump is used to keep a house warm at 22°C. How much work is required of the pump to deliver 3100 J of heat into the house if the outdoor temperature is (a) 0°C, (b) \({\bf{ - 15^\circ C}}\)? Assume a COP of 3.0. (c) Redo for both temperatures, assuming an ideal (Carnot) coefficient of performance \({\bf{COP = }}{{\bf{T}}_{\bf{L}}}{\bf{/}}\left( {{{\bf{T}}_{\bf{H}}}{\bf{ - }}{{\bf{T}}_{\bf{L}}}} \right)\).

Question: (III) The PV diagram in Fig. 15–23 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.

An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result,

(a) the temperature of the gas remains constant and the pressure decreases.

(b) both the temperature and pressure of the gas decrease.

(c) the temperature of the gas decreases and the pressure increases.

(d) both the temperature and volume of the gas increase.

(e) both the temperature and pressure of the gas increase

(III) A bowl contains many red, orange, and green jelly beans, in equal numbers. You are to make a line of 3 jelly beans by randomly taking 3 beans from the bowl.

(a) Construct a table showing the number of microstates that correspond to each macrostate. Then determine the probability of

(b) all 3 beans red, and

(c) 2 greens, 1 orange.


Question:Think up several processes (other than those already mentioned) that would obey the first law of thermodynamics, but, if they actually occurred, would violate the second law?

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