Efficiency of a heat engine whose sink is at temperate of 300 K is 40%. To increase the efficiency to 60% …

Q: Efficiency of a heat engine whose sink is at temperate of 300 K is 40%. To increase the efficiency to 60%, keeping the sink temperature constant, the source temperature must be increased by

Sol:  $\large \eta = 1 – \frac{T_2}{T_1} $

$\large \frac{T_2}{T_1} = 1 – \eta $

$\large \frac{T_2}{T_1} = 1 – \frac{40}{100} = \frac{3}{5} $

$\large T_1 = \frac{5}{3} \times T_2 = \frac{5}{3} \times 300 $

T1 = 500 K

New efficiency  η’ = 60 %

$\large \frac{T_2}{T_1′} = 1 – \eta’ $

$\large \frac{T_2}{T_1′} = 1 – \frac{60}{100} = \frac{2}{5} $

$\large T_1′ = \frac{5}{2} \times T_2 = \frac{5}{2} \times 300 $

T1‘ = 750 K

∆T = 750-500 = 250 K

A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62 K, its efficiency increases to 1/3…

Q: A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62 K, its efficiency increases to 1/3. Then T1 and T2 are, respectively .

Sol: $\large \eta_1 = 1-\frac{T_2}{T_1} $

$\large \frac{1}{6} = 1-\frac{T_2}{T_1} $

$\large \frac{T_2}{T_1} = \frac{5}{6} $ … (i)

$\large \eta_2 = 1-\frac{T_2 – 62}{T_1} $

$\large \frac{1}{3} = 1-\frac{T_2 -62}{T_1} $ …(ii)

On solving Equation (i) and (ii)

T1 = 372 K and T2 = 310 K

An ideal gas mixture filled inside a balloon expands according to the relation $PV^{2/3} = constant $ .

Q: An ideal gas mixture filled inside a balloon expands according to the relation $PV^{2/3} = constant $ . What will be the temperature inside the balloon

Sol: $\large PV^{2/3} = constant $ 

⇒ $\large  (\frac{nRT}{V}) V^{2/3} = constant $

$\large T V^{-1/3} = constant $

$\large V \propto T^3 $

Temperature increases with increase in volume.

A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20 J of work. The system is brought back…

Q: A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20 J of work. The system is brought back to its original state through a process during which 20J of heat is released by the gas. What is the work done by the gas in the second process?

Sol: dQ1 = dU1 + dW1

100 = dU1 + 20

⇒  dU1 = 80 J

dQ2 = dU2 + dW2 (∴ dU1 = -dU2)

-20 = -80 + dW2

⇒ dW2 = 60 J

In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas release 20J of heat…

Q: In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas release 20J of heat and 8J of work is done on the gas. If initial internal energy of the gas was 30J, what will be the final internal energy ?

Sol: We know that, dQ = dU + dW

Since heat is released by the system, dQ = -20 J

and work is done on the gas, dW = -8 J

dU = -20 – (-8) = -20 + 8 = -12 J

Uf – Ui = – 12 J

Uf = Ui – 12 = 30 – 12 = 18 J

A lead bullet of mass 21 g travelling at a speed of 100 ms-1 comes to rest in a wooden block…

Q: A lead bullet of mass 21 g travelling at a speed of 100 ms-1 comes to rest in a wooden block. If no heat is taken away by the wood, then find the raise in temperature of the wood. (specific heat of lead = 0.03 calorie/g°C.)

Sol: kinetic energy of the bullet = heat gained by the bullet,

$\large \frac{1}{2} m v^2 = m s \Delta T$

$\large \Delta T = \frac{v^2}{2 s}$

$\large \Delta T = \frac{(100)^2}{2 \times 0.03 \times 4.2 \times 1000}$

= 39.68°C