
A heat engine with $20\% $ efficiency does $0.100$ kJ of work during each cycle. How much heat is absorbed from the hot reservoir during each cycle?
A.$500$J
B.$400$J
C.$600$J
D.$300$J
Answer
557.1k+ views
Hint:
The efficiency of an engine is the ratio of labour done per cycle $W$ to the warmth absorbed ${Q_C}$ from the extreme temperature reservoir. It is impossible in a system for warmth transfer from a reservoir to completely convert to figure in a cyclical process within which the system returns to its initial state.
Complete step by step answer:
$W = {Q_h} - {Q_c}$
We are able to use the link between $W,{Q_{h,}},{Q_C}$ to specific the efficiency of the warmth engine in terms of ${Q_h}$ and where ${Q_{h,}}{Q_c}$ denote the warmth transfer to hot (engine) and cold (environment) reservoir.
Efficiency ($ \in $) $ = 20\% = 0.2$
$
{W}{{{Q_h}}} = {{{Q_h} - {Q_c}}}{{{Q_h}}} = 1 - {{{Q_c}}}{{{Q_h}}} \\
\\
$
Which can be written as :
${{{Q_c}}}{{{Q_h}}} = 1 - \in $
We use the definition of efficiency and the relationship between $W,{Q_{h,}}{Q_C}$ to obtain:
$ \in = {W}{{{Q_h}}} = {W}{{W + {Q_c}}}$
Dividing the numerator and denominator by W in the above equation then we get:
\[ \in = {1}{{1 + {{{Q_c}}}{W}}}\] (i)- Formula used
We know, $ \in = 0.2$
$W = 0.100$KJ
$ = 100J$
We have to find the heat released i.e. ${Q_c}$
On substituting the values in the equation above (i)
$ \in = {1}{{1 + {{{Q_c}}}{W}}} \Rightarrow 0.2 = {1}{{1 + {{{Q_C}}}{{100}}}}$
We get ${Q_c}$$ = 400J$
So the correct option is B.
Note:The second law of thermodynamics states that a Carnot engine operating between two given temperatures has the best possible efficiency of any engine operating between these two temperatures.
-Heat transfer occurs from higher temperature to lower temperature bodies but never spontaneously within the reverse direction.
-Thermal energy is the internal energy of a system in thermodynamic equilibrium because of its temperature.
-In thermodynamics, an engine could be a system that performs the conversion of warmth or thermal energy to mechanical work.
The efficiency of an engine is the ratio of labour done per cycle $W$ to the warmth absorbed ${Q_C}$ from the extreme temperature reservoir. It is impossible in a system for warmth transfer from a reservoir to completely convert to figure in a cyclical process within which the system returns to its initial state.
Complete step by step answer:
$W = {Q_h} - {Q_c}$
We are able to use the link between $W,{Q_{h,}},{Q_C}$ to specific the efficiency of the warmth engine in terms of ${Q_h}$ and where ${Q_{h,}}{Q_c}$ denote the warmth transfer to hot (engine) and cold (environment) reservoir.
Efficiency ($ \in $) $ = 20\% = 0.2$
$
{W}{{{Q_h}}} = {{{Q_h} - {Q_c}}}{{{Q_h}}} = 1 - {{{Q_c}}}{{{Q_h}}} \\
\\
$
Which can be written as :
${{{Q_c}}}{{{Q_h}}} = 1 - \in $
We use the definition of efficiency and the relationship between $W,{Q_{h,}}{Q_C}$ to obtain:
$ \in = {W}{{{Q_h}}} = {W}{{W + {Q_c}}}$
Dividing the numerator and denominator by W in the above equation then we get:
\[ \in = {1}{{1 + {{{Q_c}}}{W}}}\] (i)- Formula used
We know, $ \in = 0.2$
$W = 0.100$KJ
$ = 100J$
We have to find the heat released i.e. ${Q_c}$
On substituting the values in the equation above (i)
$ \in = {1}{{1 + {{{Q_c}}}{W}}} \Rightarrow 0.2 = {1}{{1 + {{{Q_C}}}{{100}}}}$
We get ${Q_c}$$ = 400J$
So the correct option is B.
Note:The second law of thermodynamics states that a Carnot engine operating between two given temperatures has the best possible efficiency of any engine operating between these two temperatures.
-Heat transfer occurs from higher temperature to lower temperature bodies but never spontaneously within the reverse direction.
-Thermal energy is the internal energy of a system in thermodynamic equilibrium because of its temperature.
-In thermodynamics, an engine could be a system that performs the conversion of warmth or thermal energy to mechanical work.
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