148627
A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$ absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
1 $400 \mathrm{~J}$
2 $650 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $600 \mathrm{~J}$
Explanation:
A Given, $\mathrm{T}_{1}=600 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ $\mathrm{Q}=800 \mathrm{~J}, \mathrm{~W}=?$ Efficiency of a Carnot cycle is given by, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$ $1-\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\mathrm{~W}=400 \mathrm{~J}$
Shift-II]
Thermodynamics
148628
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
1 $12 \%$
2 $24 \%$
3 $48 \%$
4 $36 \%$
Explanation:
D Let the initial efficiency of engine be $\eta$ and $\mathrm{T}_{1}$ is an absolute source temperature. When it is increased by $80 \%$ the new efficiency $=\eta+\frac{80}{100} \eta$ $=\eta+0.8 \eta=1.8 \eta$ $\therefore$ According to the question, $\eta=\frac{100-\mathrm{T}_{2}}{100}$ $1.8 \eta=\frac{125-\mathrm{T}_{2}}{125}$ From equation, (i) and (ii) We get, $\frac{\eta}{1.8 \eta}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{1}{1.8}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{5}{9}=\frac{100-\mathrm{T}_{2}}{4} \times \frac{5}{\left(125-\mathrm{T}_{2}\right)}$ $9\left(100-\mathrm{T}_{2}\right)=500-4 \mathrm{~T}_{2}$ $900-9 \mathrm{~T}_{2}=500-4 \mathrm{~T}_{2}$ $400=5 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}=\frac{400}{5}=80 \mathrm{~K}$ Then new efficiency, $\eta^{\prime}=1-\frac{T_{2}}{T_{1}}=1-\frac{80}{125}$ $=\frac{125-80}{125}=\frac{45}{125}$ $\eta^{\prime}=\frac{45}{125} \times 100=36 \%$
AP EAMCET -2016
Thermodynamics
148630
Work done by a gas in the process shown in the figure will be
1 Positive
2 Negative
3 Zero
4 Can't be determined
Explanation:
B The work of clock wise cycle is positive i.e. Turbine while the work of anti clockwise cycle is negative i.e. compressor. Hence, given cycle is anti clockwise so work done by gas is negative.
AP EAMCET-24.08.2021
Thermodynamics
148632
A Carnot engine whose heat sink is at $27^{\circ} \mathrm{C}$ has an efficiency of $40 \%$ By how much should its source temperature be changed so as to increase its efficiency to $60 \%$ ?
1 $250 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $350 \mathrm{~K}$
Explanation:
A Temperature of sink, \(\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}\) \(\qquad \eta=40 \%=0.4\) \(\qquad \eta=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.4=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}} \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}\) \(\therefore \text { New efficiency, } \eta^{\prime}=60 \%\) \(\therefore \eta^{\prime}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.6=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}^{\prime}} \Rightarrow \mathrm{T}_{\mathrm{H}}^{\prime}=750 \mathrm{~K}\) \(\text { Increase in source temperature, }\) \(\Delta \mathrm{T}_{\mathrm{H}}^{\prime}=\mathrm{T}_{\mathrm{H}}^{\prime}-\mathrm{T}_{\mathrm{H}}=750-500\) \(=250 \mathrm{~K}\)
AP EAMCET-20.08.2021
Thermodynamics
148633
A refrigerator with coefficient of performance 0.25 releases $250 \mathrm{~J}$ of heat to a hot reservoir. The work done on the working substance is
148627
A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$ absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
1 $400 \mathrm{~J}$
2 $650 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $600 \mathrm{~J}$
Explanation:
A Given, $\mathrm{T}_{1}=600 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ $\mathrm{Q}=800 \mathrm{~J}, \mathrm{~W}=?$ Efficiency of a Carnot cycle is given by, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$ $1-\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\mathrm{~W}=400 \mathrm{~J}$
Shift-II]
Thermodynamics
148628
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
1 $12 \%$
2 $24 \%$
3 $48 \%$
4 $36 \%$
Explanation:
D Let the initial efficiency of engine be $\eta$ and $\mathrm{T}_{1}$ is an absolute source temperature. When it is increased by $80 \%$ the new efficiency $=\eta+\frac{80}{100} \eta$ $=\eta+0.8 \eta=1.8 \eta$ $\therefore$ According to the question, $\eta=\frac{100-\mathrm{T}_{2}}{100}$ $1.8 \eta=\frac{125-\mathrm{T}_{2}}{125}$ From equation, (i) and (ii) We get, $\frac{\eta}{1.8 \eta}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{1}{1.8}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{5}{9}=\frac{100-\mathrm{T}_{2}}{4} \times \frac{5}{\left(125-\mathrm{T}_{2}\right)}$ $9\left(100-\mathrm{T}_{2}\right)=500-4 \mathrm{~T}_{2}$ $900-9 \mathrm{~T}_{2}=500-4 \mathrm{~T}_{2}$ $400=5 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}=\frac{400}{5}=80 \mathrm{~K}$ Then new efficiency, $\eta^{\prime}=1-\frac{T_{2}}{T_{1}}=1-\frac{80}{125}$ $=\frac{125-80}{125}=\frac{45}{125}$ $\eta^{\prime}=\frac{45}{125} \times 100=36 \%$
AP EAMCET -2016
Thermodynamics
148630
Work done by a gas in the process shown in the figure will be
1 Positive
2 Negative
3 Zero
4 Can't be determined
Explanation:
B The work of clock wise cycle is positive i.e. Turbine while the work of anti clockwise cycle is negative i.e. compressor. Hence, given cycle is anti clockwise so work done by gas is negative.
AP EAMCET-24.08.2021
Thermodynamics
148632
A Carnot engine whose heat sink is at $27^{\circ} \mathrm{C}$ has an efficiency of $40 \%$ By how much should its source temperature be changed so as to increase its efficiency to $60 \%$ ?
1 $250 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $350 \mathrm{~K}$
Explanation:
A Temperature of sink, \(\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}\) \(\qquad \eta=40 \%=0.4\) \(\qquad \eta=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.4=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}} \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}\) \(\therefore \text { New efficiency, } \eta^{\prime}=60 \%\) \(\therefore \eta^{\prime}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.6=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}^{\prime}} \Rightarrow \mathrm{T}_{\mathrm{H}}^{\prime}=750 \mathrm{~K}\) \(\text { Increase in source temperature, }\) \(\Delta \mathrm{T}_{\mathrm{H}}^{\prime}=\mathrm{T}_{\mathrm{H}}^{\prime}-\mathrm{T}_{\mathrm{H}}=750-500\) \(=250 \mathrm{~K}\)
AP EAMCET-20.08.2021
Thermodynamics
148633
A refrigerator with coefficient of performance 0.25 releases $250 \mathrm{~J}$ of heat to a hot reservoir. The work done on the working substance is
148627
A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$ absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
1 $400 \mathrm{~J}$
2 $650 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $600 \mathrm{~J}$
Explanation:
A Given, $\mathrm{T}_{1}=600 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ $\mathrm{Q}=800 \mathrm{~J}, \mathrm{~W}=?$ Efficiency of a Carnot cycle is given by, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$ $1-\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\mathrm{~W}=400 \mathrm{~J}$
Shift-II]
Thermodynamics
148628
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
1 $12 \%$
2 $24 \%$
3 $48 \%$
4 $36 \%$
Explanation:
D Let the initial efficiency of engine be $\eta$ and $\mathrm{T}_{1}$ is an absolute source temperature. When it is increased by $80 \%$ the new efficiency $=\eta+\frac{80}{100} \eta$ $=\eta+0.8 \eta=1.8 \eta$ $\therefore$ According to the question, $\eta=\frac{100-\mathrm{T}_{2}}{100}$ $1.8 \eta=\frac{125-\mathrm{T}_{2}}{125}$ From equation, (i) and (ii) We get, $\frac{\eta}{1.8 \eta}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{1}{1.8}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{5}{9}=\frac{100-\mathrm{T}_{2}}{4} \times \frac{5}{\left(125-\mathrm{T}_{2}\right)}$ $9\left(100-\mathrm{T}_{2}\right)=500-4 \mathrm{~T}_{2}$ $900-9 \mathrm{~T}_{2}=500-4 \mathrm{~T}_{2}$ $400=5 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}=\frac{400}{5}=80 \mathrm{~K}$ Then new efficiency, $\eta^{\prime}=1-\frac{T_{2}}{T_{1}}=1-\frac{80}{125}$ $=\frac{125-80}{125}=\frac{45}{125}$ $\eta^{\prime}=\frac{45}{125} \times 100=36 \%$
AP EAMCET -2016
Thermodynamics
148630
Work done by a gas in the process shown in the figure will be
1 Positive
2 Negative
3 Zero
4 Can't be determined
Explanation:
B The work of clock wise cycle is positive i.e. Turbine while the work of anti clockwise cycle is negative i.e. compressor. Hence, given cycle is anti clockwise so work done by gas is negative.
AP EAMCET-24.08.2021
Thermodynamics
148632
A Carnot engine whose heat sink is at $27^{\circ} \mathrm{C}$ has an efficiency of $40 \%$ By how much should its source temperature be changed so as to increase its efficiency to $60 \%$ ?
1 $250 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $350 \mathrm{~K}$
Explanation:
A Temperature of sink, \(\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}\) \(\qquad \eta=40 \%=0.4\) \(\qquad \eta=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.4=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}} \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}\) \(\therefore \text { New efficiency, } \eta^{\prime}=60 \%\) \(\therefore \eta^{\prime}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.6=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}^{\prime}} \Rightarrow \mathrm{T}_{\mathrm{H}}^{\prime}=750 \mathrm{~K}\) \(\text { Increase in source temperature, }\) \(\Delta \mathrm{T}_{\mathrm{H}}^{\prime}=\mathrm{T}_{\mathrm{H}}^{\prime}-\mathrm{T}_{\mathrm{H}}=750-500\) \(=250 \mathrm{~K}\)
AP EAMCET-20.08.2021
Thermodynamics
148633
A refrigerator with coefficient of performance 0.25 releases $250 \mathrm{~J}$ of heat to a hot reservoir. The work done on the working substance is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148627
A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$ absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
1 $400 \mathrm{~J}$
2 $650 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $600 \mathrm{~J}$
Explanation:
A Given, $\mathrm{T}_{1}=600 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ $\mathrm{Q}=800 \mathrm{~J}, \mathrm{~W}=?$ Efficiency of a Carnot cycle is given by, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$ $1-\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\mathrm{~W}=400 \mathrm{~J}$
Shift-II]
Thermodynamics
148628
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
1 $12 \%$
2 $24 \%$
3 $48 \%$
4 $36 \%$
Explanation:
D Let the initial efficiency of engine be $\eta$ and $\mathrm{T}_{1}$ is an absolute source temperature. When it is increased by $80 \%$ the new efficiency $=\eta+\frac{80}{100} \eta$ $=\eta+0.8 \eta=1.8 \eta$ $\therefore$ According to the question, $\eta=\frac{100-\mathrm{T}_{2}}{100}$ $1.8 \eta=\frac{125-\mathrm{T}_{2}}{125}$ From equation, (i) and (ii) We get, $\frac{\eta}{1.8 \eta}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{1}{1.8}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{5}{9}=\frac{100-\mathrm{T}_{2}}{4} \times \frac{5}{\left(125-\mathrm{T}_{2}\right)}$ $9\left(100-\mathrm{T}_{2}\right)=500-4 \mathrm{~T}_{2}$ $900-9 \mathrm{~T}_{2}=500-4 \mathrm{~T}_{2}$ $400=5 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}=\frac{400}{5}=80 \mathrm{~K}$ Then new efficiency, $\eta^{\prime}=1-\frac{T_{2}}{T_{1}}=1-\frac{80}{125}$ $=\frac{125-80}{125}=\frac{45}{125}$ $\eta^{\prime}=\frac{45}{125} \times 100=36 \%$
AP EAMCET -2016
Thermodynamics
148630
Work done by a gas in the process shown in the figure will be
1 Positive
2 Negative
3 Zero
4 Can't be determined
Explanation:
B The work of clock wise cycle is positive i.e. Turbine while the work of anti clockwise cycle is negative i.e. compressor. Hence, given cycle is anti clockwise so work done by gas is negative.
AP EAMCET-24.08.2021
Thermodynamics
148632
A Carnot engine whose heat sink is at $27^{\circ} \mathrm{C}$ has an efficiency of $40 \%$ By how much should its source temperature be changed so as to increase its efficiency to $60 \%$ ?
1 $250 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $350 \mathrm{~K}$
Explanation:
A Temperature of sink, \(\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}\) \(\qquad \eta=40 \%=0.4\) \(\qquad \eta=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.4=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}} \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}\) \(\therefore \text { New efficiency, } \eta^{\prime}=60 \%\) \(\therefore \eta^{\prime}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.6=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}^{\prime}} \Rightarrow \mathrm{T}_{\mathrm{H}}^{\prime}=750 \mathrm{~K}\) \(\text { Increase in source temperature, }\) \(\Delta \mathrm{T}_{\mathrm{H}}^{\prime}=\mathrm{T}_{\mathrm{H}}^{\prime}-\mathrm{T}_{\mathrm{H}}=750-500\) \(=250 \mathrm{~K}\)
AP EAMCET-20.08.2021
Thermodynamics
148633
A refrigerator with coefficient of performance 0.25 releases $250 \mathrm{~J}$ of heat to a hot reservoir. The work done on the working substance is
148627
A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$ absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
1 $400 \mathrm{~J}$
2 $650 \mathrm{~J}$
3 $750 \mathrm{~J}$
4 $600 \mathrm{~J}$
Explanation:
A Given, $\mathrm{T}_{1}=600 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ $\mathrm{Q}=800 \mathrm{~J}, \mathrm{~W}=?$ Efficiency of a Carnot cycle is given by, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$ $1-\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\frac{300}{600}=\frac{\mathrm{W}}{800}$ $\mathrm{~W}=400 \mathrm{~J}$
Shift-II]
Thermodynamics
148628
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
1 $12 \%$
2 $24 \%$
3 $48 \%$
4 $36 \%$
Explanation:
D Let the initial efficiency of engine be $\eta$ and $\mathrm{T}_{1}$ is an absolute source temperature. When it is increased by $80 \%$ the new efficiency $=\eta+\frac{80}{100} \eta$ $=\eta+0.8 \eta=1.8 \eta$ $\therefore$ According to the question, $\eta=\frac{100-\mathrm{T}_{2}}{100}$ $1.8 \eta=\frac{125-\mathrm{T}_{2}}{125}$ From equation, (i) and (ii) We get, $\frac{\eta}{1.8 \eta}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{1}{1.8}=\frac{100-\mathrm{T}_{2}}{100} \times \frac{125}{125-\mathrm{T}_{2}}$ $\frac{5}{9}=\frac{100-\mathrm{T}_{2}}{4} \times \frac{5}{\left(125-\mathrm{T}_{2}\right)}$ $9\left(100-\mathrm{T}_{2}\right)=500-4 \mathrm{~T}_{2}$ $900-9 \mathrm{~T}_{2}=500-4 \mathrm{~T}_{2}$ $400=5 \mathrm{~T}_{2}$ $\mathrm{~T}_{2}=\frac{400}{5}=80 \mathrm{~K}$ Then new efficiency, $\eta^{\prime}=1-\frac{T_{2}}{T_{1}}=1-\frac{80}{125}$ $=\frac{125-80}{125}=\frac{45}{125}$ $\eta^{\prime}=\frac{45}{125} \times 100=36 \%$
AP EAMCET -2016
Thermodynamics
148630
Work done by a gas in the process shown in the figure will be
1 Positive
2 Negative
3 Zero
4 Can't be determined
Explanation:
B The work of clock wise cycle is positive i.e. Turbine while the work of anti clockwise cycle is negative i.e. compressor. Hence, given cycle is anti clockwise so work done by gas is negative.
AP EAMCET-24.08.2021
Thermodynamics
148632
A Carnot engine whose heat sink is at $27^{\circ} \mathrm{C}$ has an efficiency of $40 \%$ By how much should its source temperature be changed so as to increase its efficiency to $60 \%$ ?
1 $250 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $350 \mathrm{~K}$
Explanation:
A Temperature of sink, \(\mathrm{T}_{\mathrm{L}}=300 \mathrm{~K}\) \(\qquad \eta=40 \%=0.4\) \(\qquad \eta=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.4=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}} \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}\) \(\therefore \text { New efficiency, } \eta^{\prime}=60 \%\) \(\therefore \eta^{\prime}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) \(0.6=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}^{\prime}} \Rightarrow \mathrm{T}_{\mathrm{H}}^{\prime}=750 \mathrm{~K}\) \(\text { Increase in source temperature, }\) \(\Delta \mathrm{T}_{\mathrm{H}}^{\prime}=\mathrm{T}_{\mathrm{H}}^{\prime}-\mathrm{T}_{\mathrm{H}}=750-500\) \(=250 \mathrm{~K}\)
AP EAMCET-20.08.2021
Thermodynamics
148633
A refrigerator with coefficient of performance 0.25 releases $250 \mathrm{~J}$ of heat to a hot reservoir. The work done on the working substance is
