NEET-XII-Physics
26: Laws of Thermodynamics
- #2-aHow 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.
FigureAns : 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.
`` \Rightarrow ``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.
`` \Rightarrow `` 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\,\mathrm{\,K\,}``
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- #2-bHow much work is done on the liquid?Ans : 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.
`` \Rightarrow ``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
- #2-cCalculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.
FigureAns : Suppose ∆t is the rise in temperature of the paddle wheel when the system gains energy.
`` \Rightarrow `` 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\,\mathrm{\,K\,}``
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