A carnot engine having an efficiency of $1/10$ as heat engine, is used as a refrigerator. If the work done on the system is $10\text{ J}$, the amount of energy absorbed from the reservoir at lower temperature is:
(2017-Delhi)
Coefficient of performance $\beta = \frac{1-eta}{\eta} = 9$. Heat absorbed from lower reservoir $Q_2 = \beta \times W = 9 \times 10\text{ J} = 90\text{ J}$.
Carnot engine, having an efficiency of $\eta = 1/10$. As heat engine, is used as a refrigerator. If the work done on the system is $10\text{ J}$, the amount of energy absorbed from the reservoir at lower temperature is: (2015)
Using $\beta = \frac{1-\eta}{\eta} = 9$, the heat absorbed at lower temperature is $Q_2 = \beta W = 9 \times 10\text{ J} = 90\text{ J}$.
The efficiency of Carnot engine is $50\%$ and temperature of sink is $500\text{ K}$. If temperature of source is kept constant and its efficiency raised to $60\%$, then the required temperature of the sink will be: (2007)
Source temperature $T_1 = \frac{T_2}{1-\eta_1} = \frac{500}{0.5} = 1000\text{ K}$. New sink temperature $T_2' = T_1(1-\eta_2) = 1000(1 - 0.6) = 400\text{ K}$.
An engine has an efficiency of $1/6$. When the temperature of sink is reduced by $62\text{ }^\circ\text{C}$, its efficiency is doubled. Temperature of the source is (2007)
Initial efficiency $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{1}{6}$. After reduction, $\eta_2 = 2(\frac{1}{6}) = \frac{1}{3} = 1 - \frac{T_2-62}{T_1}$. Solving these equations yields source temperature $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$.
A Carnot engine whose sink is at $300\text{ K}$ 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? (2006)
Initial source temperature $T_1 = \frac{300}{1-0.4} = 500\text{ K}$. New efficiency $\eta' = 1.5 \times 0.4 = 0.6$. New source temperature $T_1' = \frac{300}{1-0.6} = 750\text{ K}$. Increase $\Delta T = 750 - 500 = 250\text{ K}$.
An ideal gas heat engine operates in Carnot cycle between $227\text{ }^\circ\text{C}$ and $127\text{ }^\circ\text{C}$. It absorbs $6 \times 10^4\text{ cal}$ of heat at higher temperature. Amount of heat converted to work is: (2005)
Temperatures are $T_1 = 500\text{ K}$ and $T_2 = 400\text{ K}$. Efficiency $\eta = 1 - \frac{400}{500} = 0.2$. Work $W = \eta Q_1 = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4\text{ cal}$.
An ideal gas heat engine operates in a Carnot cycle. Between $227\text{ }^\circ\text{C}$ and $127\text{ }^\circ\text{C}$. It absorbs $6\text{ kcal}$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to: (2003)
Efficiency $\eta = 1 - \frac{400}{500} = 0.2$. Work done $W = \eta Q_1 = 0.2 \times 6\text{ kcal} = 1.2\text{ kcal}$.
The efficiency of carnot engine is $50\%$ and temperature of sink is $500\text{ K}$. If temperature of source is kept constant and its efficiency raised to $60\%$, then the required temperature of the sink will be: (2002)
Source temperature $T_1 = \frac{500}{1-0.5} = 1000\text{ K}$. New sink temperature $T_2' = T_1(1-\eta_2) = 1000(1-0.6) = 400\text{ K}$.
A scientist says that the efficiency of his heat engine which work at source temperature $127\text{ }^\circ\text{C}$ and sink temperature $27\text{ }^\circ\text{C}$ is $26\%$, then: (2001)
Maximum Carnot efficiency $\eta_{\text{max}} = 1 - \frac{300}{400} = 25\%$. Since the claimed efficiency ($26\%$) exceeds the Carnot limit, it is impossible.
The ratio ($W/Q$) for a carnot-engine is $1/6$. Now the temperature of sink is reduced by $62\text{ }^\circ\text{C}$, this ratio becomes twice, therefore the initial temp. of the sink and source are respectively: (2000)
Ratio $W/Q_1$ is the efficiency $\eta_1 = 1/6$. When sink temperature decreases, efficiency becomes $2/6 = 1/3$. Solving yields source $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$ and sink $T_2 = 310\text{ K} = 37\text{ }^\circ\text{C}$.