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Question:(II) A heat engine uses a heat source at 580°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?

Short Answer

Expert verified

The temperature of heat source is \(1147.06\;{\rm{K}}\).

Step by step solution

01

Given Data

The temperature of engine is \({T_1} = 580^\circ {\rm{C}}\).

The Carnot efficiency is \(\eta = 22\% \).

The ideal efficiency is \({\eta _{\rm{i}}} = 42\% \).

02

Understanding Carnot efficiency

In this problem, use the Carnot efficiency to find the temperature at exhaust. Again, recalculate the temperature of the heat source with the help of new ideal efficiency.

03

Calculation of exhaust temperature of engine

The relation of Carnot efficiency is given by,

\(\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right)\)

Here, \({T_2}\) is the temperature at exhaust.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.22 &= \left( {1 - \frac{{{T_2}}}{{\left( {580^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\{T_2} &= 665.3\;{\rm{K}}\end{aligned}\)

04

Calculation of temperature of the heat source

The relation to find temperature is given by,

\({\eta _{\rm{i}}} = \left( {1 - \frac{{{T_2}}}{{{{T'}_1}}}} \right)\)

Here, \({T'_1}\) is the temperature of heat source.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.42 &= \left( {1 - \frac{{665.3\;{\rm{K}}}}{{{T_1}^\prime }}} \right)\\{T_1}^\prime &= 1147.06\;{\rm{K}}\end{aligned}\)

Thus, \({T_1}^\prime = 1147.06\;{\rm{K}}\) is the required rate of heat output.

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Most popular questions from this chapter

(II) A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from to \({\bf{16}}{\bf{.2}}\;{{\bf{m}}^{\bf{3}}}\).Calculate (a) the work done by the gas and (b) the change in internal energy of the gas.

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.

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


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?

Question: An ideal gas undergoes an isothermal expansion from state A to state B. In this process (use sign conventions, page 413),

(a) \[Q = 0,\;\Delta U = 0,\;W > 0\].

(b) \[Q > 0,\;\Delta U = 0,\;W < 0\].

(c) \[Q = 0,\;\Delta U > 0,\;W > 0\].

(d) \[Q > 0,\;\Delta U = 0,W > 0\].

(e) \[Q = 0,\;\Delta U < 0,\;W < 0\].

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