Second Law Of Thermodynamics NAT Level - 2 - Physics MCQ

To obtain the heat that is delivered to the room, first we have to find out the required work W to run the freezer.
To obtain the required work W to run the freezer,

Now 
⇒ heat delivered to the room would be 217.5kJ
The correct answer is: 217.5

There is no change in the entropy of the resistance because there is no change in the state of the wire.
The correct answer is: 0

Hence, m₁ = ?, m₂ = 1500gm
T₁ = 373K, T₂ = 273 + 15 = 288 K
Final temperature of mixture be T = 353 K
m₁ × 540 + m₁ (373 – 353) = 1500(353 – 288)
(540 + 20)m₁ = 1500 × 65
m₁ = 175g
The correct answer is: 175

W = Q_K – Q_L

= 20/30
K = 0.67
The correct answer is: 0.67

Since the rod is thermally insulated from the surrounding and its two ends are immersed in two heat reservoirs. then,
 where K , is thermal conductivity.
Now, increase in entropy is given by



T₁ = 800K, T₂ = 500K

The correct answer is: 0.225

Now,

Also, the higher temperature loses heat and water temperature gains heat, we can write,



Since 
and  
Total entropy change of universe




m = 1000gm, C_P = 0.01Cal K^-1 gm^-1
T₁ = 300K, T₂ = 500K
⇒ ΔS = 0.645
The correct answer is: 0.645

Let work done by the engine B is W_B.

Let heat input of engine B is Q_i,B.
⇒ 
Let heat output of engine B is Q_0,B
⇒ 
Work done in an engine is equal to the difference of input heat and output heat.


⇒ Efficiency of engine B will be

The correct answer is: 0.33

S = K_blnΩ
At 0 K, we know the system is certainly in the ground state Ω = 1, so the definition returns zero.
The correct answer is: 0

Mass of water = 1.78kg
Mass of ice = 262g
Mass of ice-water mixture = 1.78 kg + 262g
= 2.04kg
If eventually, ice and and water have the same mass, then the final state will have 1.02kg of each.
Thus, mass of water that changed into ice m will be the difference of mass of water m_W and mass of final state m_S .
So, m = m_W – m_S
⇒m = 1.78 – 1.02
= 0.76kg
The change of water at 0ºC to ice at 0ºC is isothermal.
⇒ ΔS = –mL/T
= –(0.76kg) (333×103J/kg) (273K)
= –927 J/K
⇒ Change in entropy of the system during this process will be –927J/K.
The correct answer is: 927

Change in entropy = 

= 0.33Js^-1K^-1
The correct answer is: 0.33