Q. 2

# 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 JK–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.

Given

Mass attaches to the pulley m = 12kg

Heat capacity of liquid s =4200 JK-1

Height through which mass fall= 70cm=0.7m

a) Paddle immersed in liquid is kept in an adiabatic container. So, no heat can be either supplied or extracted to the liquid. Therefore, heat given to liquid is zero.

b) As no heat is supplied to liquid and pulley is frictionless, work done on the liquid will be equal to the potential energy of the mass attached to the pulley.

Work done on the liquid = potential energy of mass

Potential energy of mass= mgh

Where g=acceleration due to gravity=10ms-2

Potential energy of mass= 12×10×0.7=84J

work done on liquid =84J.

c) The mechanical work calculated in the second part will be converted into heat. This heat will be supplied to liquid and due to which temperature of the liquid will rise.

We know that,

Heat capacity

Where ΔQ = heat supplied

ΔT=rise in temperature

Since work done is equal to heat supplied, therefore

the rise in temperature of the liquid will be 0.02K.

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