A sample of 0.1g of water at 100°C and normal pressure (1.013 x 105 Nm-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is -

Correct option is C

The correct answer is (c) 208.7 joule. To calculate the change in internal energy (ΔU) of the sample, we use the first law of thermodynamics: ΔU = Q - W, where Q is the heat added to the system, and W is the work done by the system.
· Now, it is given that-
· Pressure (P) = 1.013 × 105 Nm-2
· Final volume (V2) = 167.1 cc = 167.1 × 10-6 m3 (since 1 cc = 10-6 m3)
· Since initially there is water, so Initial volume (V1) = 0
· Work done (Δ W) = P (V2 - V1)
· = 1.013 × 105 Nm-2× (167.1 × 10-6 m3- 0) = 16.93 J
· Heat exchanged (Δ Q) = 54 cal = 54 × 4.18 J = 225.72 J (Since 1 Cal = 4.18 J)
· Δ Q = Δ U + Δ W
· Change in internal energy (Δ U) = Δ Q - Δ W = 225.72 - 16.93 = 208.7 Joule.
Important Key Points:
· The change in internal energy for the conversion of water to steam at 100°C is approximately 208.7 J.
· This calculation uses the first law of thermodynamics: ΔU = Q - W.
· The specific heat capacity of water and the work done by expanding steam are crucial for the calculation.
· This energy change represents the internal energy difference between the water and steam states.
· Understanding this concept is vital for thermodynamics and energy conversion studies.