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प्रश्न
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 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.
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उत्तर
(a) Heat is not given to the liquid; instead, the mechanical work done is converted to heat. Also the container is adiabatic. So, no heat can enter or exit the system. This implies that the heat given to the liquid is zero.
(b) Since the 12 kg mass falls through a distance of 70 cm under gravity, energy is lost by this mass. As this mass is connected to the paddle wheel, energy lost by this mass is gained by the paddle wheel.
⇒">⇒ Work done on the liquid = PE lost by the 12 kg mass
Now,
PE lost by the 12 kg mass = mgh
= 12 × 10 × 0.70
= 84 J
(c) Suppose ∆t is the rise in temperature of the paddle wheel when the system gains energy.
⇒">⇒ 84 = ms∆t
If s is the specific heat of the system, then
84 = 1 × 4200 × ∆t .........(for 'm' = 1 kg)
\[∆ t = \frac{84}{4200} = \frac{1}{50} = 0 . 02 K\]
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