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.
Figure
Answers
The rise in the temperature of the liquid is 0.02 K
Explanation:
(a) The container is adiabatic therefore 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 y this mass.
Work done on the liquid = P.E lost by the 12 kg mass
Now
P.E lost by the 12 kg mass = mgh
= 12 x 10 x 0.70 = 84 J
(c) Suppose Δt is the rise in temp. of the paddle wheel when system gains energy.
84 = msΔt
If s is the specific heat of the system then
84 = 1 x 4200 x Δt
Δt = 84 / 4200 = 1/50 = 0.02 K
The rise in the temperature of the liquid is 0.02 K
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(a) Heat supplied to the liquid is null.
(b) Work done on the liquid is 84 J.
(c) The rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle is 0.02 K.
Explanation:
(a) The liquid does not absorb power; rather, the mechanical work done is transformed into heat. The container itself is adiabatic, too. So, the device cannot enter or exit any heat. This implies the heat supplied to the liquid is null.
(b) Because the 12 kg mass is under gravity through a distance of 70 cm, this mass loses energy.
As this mass is related to the paddle wheel, the paddle wheel recovers the energy lost by that mass.
Work done on liquid = Potential energy lost by the mass of 12 kg
Potential energy lost by the mass of 12 kg = mgh
= 12 × 10 × 0.70
= 120 × 0.70
= 84 J
(c) Suppose ∆t is the paddle wheel's temperature rise when the machine absorbs energy.84 = ms∆t
System specific heat
m = 1 kg
84 = 1 × 4200 × ∆t