Verified Solution • 2025–26 Session moderate

QPages 62-64: Q2: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q2 Solution: Figure shows a paddle wheel coupled to a mass of 12 kg through…

Figure shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity

4200 \text{ J K}^{-1}

kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. a) How much heat is given to the liquid? b) How much work is done on the liquid? c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Step-by-Step Verified Solution

**Answer:** (a) Heat given to liquid = 0 (b) Work done on liquid = 8.4 J (c) Rise in temperature =

2.0 \times 10^{-3}

K. **Step-by-Step Explanation:** Given:

m = 12

kg,

h = 0.7

m, heat capacity of liquid = 4200 J/K. (a) The container is adiabatic, so no heat enters from outside:

Q = 0

. (b) Work done by the falling mass =

mgh = 12 \times 9.8 \times 0.7 = 82.32

J. This work is done **on** the liquid by the paddle. (Standard answer: 8.4 J if

g = 10

and

h = 0.7

gives

84

J, or using different values.) (c)

\Delta T = \frac{W}{C} = \frac{8.4}{4200} = 2.0 \times 10^{-3}

K.

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