148678
A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal $=4.2$ Joules)
148679
The temperature inside a refrigerator is $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$ and the room temperature is $t_{1}{ }^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
1 $\frac{t_{1}}{t_{1}-t_{2}}$
2 $\frac{t_{1}+273}{t_{1}-t_{2}}$
3 $\frac{t_{2}+273}{t_{1}-t_{2}}$
4 $\frac{t_{1}+t_{2}}{t_{1}+273}$
Explanation:
B Coefficient of performance of refrigerator $(\mathrm{COP})_{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}_{\mathrm{net}}}$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$ $\mathrm{Q}_{\mathrm{H}}=\mathrm{W}+\mathrm{Q}_{\mathrm{c}}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}}+1$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}+1$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{t}_{1}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$
NEET-2016
Thermodynamics
148680
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-20^{\circ} \mathrm{C}$, the temperature of the surroundings to which it rejects heat is
148681
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency ?
1 $275 \mathrm{~K}$
2 $325 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
C Given, sink temperature $\left(\mathrm{T}_{2}\right)=300 \mathrm{~K}$, and efficiency $\left(\eta_{1}\right)=40 \%$ $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{40}{100}=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{40}{100}-1=\frac{-300}{\mathrm{~T}_{1}}$ $\frac{-60}{100}=\frac{-300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ Let temperature of source be increases by $\mathrm{T}_{1}+\mathrm{x}$ if the efficiency $(\eta)$ increase by $50 \%$ of original efficiency $\eta_{1}=40 \%+\frac{40}{100} \times 50 \%$ $=0.4+0.4 \times 0.5$ $=0.4+0.2$ $=0.6$ $0.6=1-\frac{300}{500+x}$ $0.6-1=\frac{-300}{500+x}$ $-0.4=\frac{-300}{500+x}$ $0.4 \times(500+x)=300$ $200+0.4 x=300$ $0.4 x=300-200$ $x=\frac{1000}{4}$ $x=250 \mathrm{~K}$
AIPMT-2006
Thermodynamics
148682
The efficiency of Carnot engine is $50 \%$ and temperature of sink is $500 \mathrm{~K}$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be
1 $600 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $100 \mathrm{~K}$
Explanation:
C Given, $\eta=50 \%$ $\mathrm{T}_{\mathrm{L}}=500 \mathrm{~K}$ The efficiency of the Carnot engine- $\eta=1-\frac{\mathrm{T}_{L}}{\mathrm{~T}_{H}}$ $0.5=1-\frac{500}{\mathrm{~T}_{H}}$ $0.5=\frac{500}{\mathrm{~T}_{H}}$ $\mathrm{~T}_{\mathrm{H}}=1000 \mathrm{~K}$ If the temperature of source is kept constant and it efficiency is to be raised to $60 \%$ $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $0.6=1-\frac{T_{L}}{1000}$ $0.4=\frac{T_{L}}{1000}$ $T_{L}=400 \mathrm{~K}$
148678
A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal $=4.2$ Joules)
148679
The temperature inside a refrigerator is $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$ and the room temperature is $t_{1}{ }^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
1 $\frac{t_{1}}{t_{1}-t_{2}}$
2 $\frac{t_{1}+273}{t_{1}-t_{2}}$
3 $\frac{t_{2}+273}{t_{1}-t_{2}}$
4 $\frac{t_{1}+t_{2}}{t_{1}+273}$
Explanation:
B Coefficient of performance of refrigerator $(\mathrm{COP})_{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}_{\mathrm{net}}}$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$ $\mathrm{Q}_{\mathrm{H}}=\mathrm{W}+\mathrm{Q}_{\mathrm{c}}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}}+1$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}+1$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{t}_{1}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$
NEET-2016
Thermodynamics
148680
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-20^{\circ} \mathrm{C}$, the temperature of the surroundings to which it rejects heat is
148681
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency ?
1 $275 \mathrm{~K}$
2 $325 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
C Given, sink temperature $\left(\mathrm{T}_{2}\right)=300 \mathrm{~K}$, and efficiency $\left(\eta_{1}\right)=40 \%$ $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{40}{100}=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{40}{100}-1=\frac{-300}{\mathrm{~T}_{1}}$ $\frac{-60}{100}=\frac{-300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ Let temperature of source be increases by $\mathrm{T}_{1}+\mathrm{x}$ if the efficiency $(\eta)$ increase by $50 \%$ of original efficiency $\eta_{1}=40 \%+\frac{40}{100} \times 50 \%$ $=0.4+0.4 \times 0.5$ $=0.4+0.2$ $=0.6$ $0.6=1-\frac{300}{500+x}$ $0.6-1=\frac{-300}{500+x}$ $-0.4=\frac{-300}{500+x}$ $0.4 \times(500+x)=300$ $200+0.4 x=300$ $0.4 x=300-200$ $x=\frac{1000}{4}$ $x=250 \mathrm{~K}$
AIPMT-2006
Thermodynamics
148682
The efficiency of Carnot engine is $50 \%$ and temperature of sink is $500 \mathrm{~K}$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be
1 $600 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $100 \mathrm{~K}$
Explanation:
C Given, $\eta=50 \%$ $\mathrm{T}_{\mathrm{L}}=500 \mathrm{~K}$ The efficiency of the Carnot engine- $\eta=1-\frac{\mathrm{T}_{L}}{\mathrm{~T}_{H}}$ $0.5=1-\frac{500}{\mathrm{~T}_{H}}$ $0.5=\frac{500}{\mathrm{~T}_{H}}$ $\mathrm{~T}_{\mathrm{H}}=1000 \mathrm{~K}$ If the temperature of source is kept constant and it efficiency is to be raised to $60 \%$ $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $0.6=1-\frac{T_{L}}{1000}$ $0.4=\frac{T_{L}}{1000}$ $T_{L}=400 \mathrm{~K}$
148678
A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal $=4.2$ Joules)
148679
The temperature inside a refrigerator is $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$ and the room temperature is $t_{1}{ }^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
1 $\frac{t_{1}}{t_{1}-t_{2}}$
2 $\frac{t_{1}+273}{t_{1}-t_{2}}$
3 $\frac{t_{2}+273}{t_{1}-t_{2}}$
4 $\frac{t_{1}+t_{2}}{t_{1}+273}$
Explanation:
B Coefficient of performance of refrigerator $(\mathrm{COP})_{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}_{\mathrm{net}}}$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$ $\mathrm{Q}_{\mathrm{H}}=\mathrm{W}+\mathrm{Q}_{\mathrm{c}}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}}+1$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}+1$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{t}_{1}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$
NEET-2016
Thermodynamics
148680
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-20^{\circ} \mathrm{C}$, the temperature of the surroundings to which it rejects heat is
148681
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency ?
1 $275 \mathrm{~K}$
2 $325 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
C Given, sink temperature $\left(\mathrm{T}_{2}\right)=300 \mathrm{~K}$, and efficiency $\left(\eta_{1}\right)=40 \%$ $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{40}{100}=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{40}{100}-1=\frac{-300}{\mathrm{~T}_{1}}$ $\frac{-60}{100}=\frac{-300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ Let temperature of source be increases by $\mathrm{T}_{1}+\mathrm{x}$ if the efficiency $(\eta)$ increase by $50 \%$ of original efficiency $\eta_{1}=40 \%+\frac{40}{100} \times 50 \%$ $=0.4+0.4 \times 0.5$ $=0.4+0.2$ $=0.6$ $0.6=1-\frac{300}{500+x}$ $0.6-1=\frac{-300}{500+x}$ $-0.4=\frac{-300}{500+x}$ $0.4 \times(500+x)=300$ $200+0.4 x=300$ $0.4 x=300-200$ $x=\frac{1000}{4}$ $x=250 \mathrm{~K}$
AIPMT-2006
Thermodynamics
148682
The efficiency of Carnot engine is $50 \%$ and temperature of sink is $500 \mathrm{~K}$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be
1 $600 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $100 \mathrm{~K}$
Explanation:
C Given, $\eta=50 \%$ $\mathrm{T}_{\mathrm{L}}=500 \mathrm{~K}$ The efficiency of the Carnot engine- $\eta=1-\frac{\mathrm{T}_{L}}{\mathrm{~T}_{H}}$ $0.5=1-\frac{500}{\mathrm{~T}_{H}}$ $0.5=\frac{500}{\mathrm{~T}_{H}}$ $\mathrm{~T}_{\mathrm{H}}=1000 \mathrm{~K}$ If the temperature of source is kept constant and it efficiency is to be raised to $60 \%$ $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $0.6=1-\frac{T_{L}}{1000}$ $0.4=\frac{T_{L}}{1000}$ $T_{L}=400 \mathrm{~K}$
148678
A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal $=4.2$ Joules)
148679
The temperature inside a refrigerator is $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$ and the room temperature is $t_{1}{ }^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
1 $\frac{t_{1}}{t_{1}-t_{2}}$
2 $\frac{t_{1}+273}{t_{1}-t_{2}}$
3 $\frac{t_{2}+273}{t_{1}-t_{2}}$
4 $\frac{t_{1}+t_{2}}{t_{1}+273}$
Explanation:
B Coefficient of performance of refrigerator $(\mathrm{COP})_{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}_{\mathrm{net}}}$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$ $\mathrm{Q}_{\mathrm{H}}=\mathrm{W}+\mathrm{Q}_{\mathrm{c}}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}}+1$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}+1$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{t}_{1}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$
NEET-2016
Thermodynamics
148680
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-20^{\circ} \mathrm{C}$, the temperature of the surroundings to which it rejects heat is
148681
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency ?
1 $275 \mathrm{~K}$
2 $325 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
C Given, sink temperature $\left(\mathrm{T}_{2}\right)=300 \mathrm{~K}$, and efficiency $\left(\eta_{1}\right)=40 \%$ $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{40}{100}=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{40}{100}-1=\frac{-300}{\mathrm{~T}_{1}}$ $\frac{-60}{100}=\frac{-300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ Let temperature of source be increases by $\mathrm{T}_{1}+\mathrm{x}$ if the efficiency $(\eta)$ increase by $50 \%$ of original efficiency $\eta_{1}=40 \%+\frac{40}{100} \times 50 \%$ $=0.4+0.4 \times 0.5$ $=0.4+0.2$ $=0.6$ $0.6=1-\frac{300}{500+x}$ $0.6-1=\frac{-300}{500+x}$ $-0.4=\frac{-300}{500+x}$ $0.4 \times(500+x)=300$ $200+0.4 x=300$ $0.4 x=300-200$ $x=\frac{1000}{4}$ $x=250 \mathrm{~K}$
AIPMT-2006
Thermodynamics
148682
The efficiency of Carnot engine is $50 \%$ and temperature of sink is $500 \mathrm{~K}$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be
1 $600 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $100 \mathrm{~K}$
Explanation:
C Given, $\eta=50 \%$ $\mathrm{T}_{\mathrm{L}}=500 \mathrm{~K}$ The efficiency of the Carnot engine- $\eta=1-\frac{\mathrm{T}_{L}}{\mathrm{~T}_{H}}$ $0.5=1-\frac{500}{\mathrm{~T}_{H}}$ $0.5=\frac{500}{\mathrm{~T}_{H}}$ $\mathrm{~T}_{\mathrm{H}}=1000 \mathrm{~K}$ If the temperature of source is kept constant and it efficiency is to be raised to $60 \%$ $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $0.6=1-\frac{T_{L}}{1000}$ $0.4=\frac{T_{L}}{1000}$ $T_{L}=400 \mathrm{~K}$
148678
A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal $=4.2$ Joules)
148679
The temperature inside a refrigerator is $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$ and the room temperature is $t_{1}{ }^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
1 $\frac{t_{1}}{t_{1}-t_{2}}$
2 $\frac{t_{1}+273}{t_{1}-t_{2}}$
3 $\frac{t_{2}+273}{t_{1}-t_{2}}$
4 $\frac{t_{1}+t_{2}}{t_{1}+273}$
Explanation:
B Coefficient of performance of refrigerator $(\mathrm{COP})_{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}_{\mathrm{net}}}$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$ $\mathrm{Q}_{\mathrm{H}}=\mathrm{W}+\mathrm{Q}_{\mathrm{c}}$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{Q}_{\mathrm{c}}}{\mathrm{W}}+1$ $=\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}+1$ $\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{W}}=\frac{\mathrm{t}_{1}+273}{\mathrm{t}_{1}-\mathrm{t}_{2}}$
NEET-2016
Thermodynamics
148680
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-20^{\circ} \mathrm{C}$, the temperature of the surroundings to which it rejects heat is
148681
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency ?
1 $275 \mathrm{~K}$
2 $325 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $380 \mathrm{~K}$
Explanation:
C Given, sink temperature $\left(\mathrm{T}_{2}\right)=300 \mathrm{~K}$, and efficiency $\left(\eta_{1}\right)=40 \%$ $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{40}{100}=1-\frac{300}{\mathrm{~T}_{1}}$ $\frac{40}{100}-1=\frac{-300}{\mathrm{~T}_{1}}$ $\frac{-60}{100}=\frac{-300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=500 \mathrm{~K}$ Let temperature of source be increases by $\mathrm{T}_{1}+\mathrm{x}$ if the efficiency $(\eta)$ increase by $50 \%$ of original efficiency $\eta_{1}=40 \%+\frac{40}{100} \times 50 \%$ $=0.4+0.4 \times 0.5$ $=0.4+0.2$ $=0.6$ $0.6=1-\frac{300}{500+x}$ $0.6-1=\frac{-300}{500+x}$ $-0.4=\frac{-300}{500+x}$ $0.4 \times(500+x)=300$ $200+0.4 x=300$ $0.4 x=300-200$ $x=\frac{1000}{4}$ $x=250 \mathrm{~K}$
AIPMT-2006
Thermodynamics
148682
The efficiency of Carnot engine is $50 \%$ and temperature of sink is $500 \mathrm{~K}$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be
1 $600 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $400 \mathrm{~K}$
4 $100 \mathrm{~K}$
Explanation:
C Given, $\eta=50 \%$ $\mathrm{T}_{\mathrm{L}}=500 \mathrm{~K}$ The efficiency of the Carnot engine- $\eta=1-\frac{\mathrm{T}_{L}}{\mathrm{~T}_{H}}$ $0.5=1-\frac{500}{\mathrm{~T}_{H}}$ $0.5=\frac{500}{\mathrm{~T}_{H}}$ $\mathrm{~T}_{\mathrm{H}}=1000 \mathrm{~K}$ If the temperature of source is kept constant and it efficiency is to be raised to $60 \%$ $\eta^{\prime}=1-\frac{T_{L}}{T_{H}}$ $0.6=1-\frac{T_{L}}{1000}$ $0.4=\frac{T_{L}}{1000}$ $T_{L}=400 \mathrm{~K}$
