Question 381:
moderateA 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.
Question 382:
moderateThe 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 $T_2 = 310\text{ K} = 37\text{ }^\circ\text{C}$.
Question 383:
moderateThe efficiency of a Carnot engine operating with reservoir temperature of $100\text{ }^\circ\text{C}$ and $-23\text{ }^\circ\text{C}$ will be: (1997)
Efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{250\text{ K}}{373\text{ K}} = \frac{373-250}{373}$.
Question 384:
moderateAn ideal Carnot engine, whose efficiency is $40\%$, receives heat at $500\text{ K}$. If its efficiency is $50\%$, the intake temperature for the same exhaust temperature is: (1995)
Exhaust temperature $T_2 = 500(1-0.4) = 300\text{ K}$. New intake temperature $T_1' = \frac{300}{1-0.5} = 600\text{ K}$.
Question 385:
moderateA 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 $Q_2 = \beta \times W = 9 \times 10\text{ J} = 90\text{ J}$.
Question 386:
moderateCarnot 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 $Q_2 = \beta W = 9 \times 10\text{ J} = 90\text{ J}$.
Question 387:
moderateThe 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 $T_2' = T_1(1-\eta_2) = 1000(1 - 0.6) = 400\text{ K}$.
Question 388:
moderateAn 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 $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$.
Question 389:
easyThe value of critical temperature in terms of Van der Waals’ constant $a$ and $b$ is given by: (1996)
The critical temperature $T_c$ for a real gas obeying the Van der Waals equation is theoretically derived as $T_c = \frac{8a}{27Rb}$, where $a$ and $b$ are the Van der Waals constants.
Question 390:
easyAccording to kinetic theory of gases, at absolute zero temperature (1990)
Kinetic energy is directly proportional to the absolute temperature ($E_k = \frac{3}{2}kT$). At absolute zero ($T = 0 \text{ K}$), the translational kinetic energy becomes zero, implying that all molecular motion stops.