I don\'t understand how to solve this can someone provide me with a step by step

Question

I don't understand how to solve this can someone provide me with a step by step solution

P4.10 A Joule-type paddlewheel is fitted with two weights, one with mass M1 = 150 kg,
and the other with a variable mass M2. As they descend slowly under gravity, they
both do work on the water. The container holds 5 kg of water.

(a) If only the first mass M1 is attached, and it slowly descends a distance d = 5
m, how much work would be done, and what would be the temperature change of
the water?

(b) If the second mass also equaled 150 kg, and it was also attached, answer the
same question.

(c) If the second mass equaled 75 kg, and it was also attached, answer the same
question.

Explanation / Answer

a) Workdone on water = decrease in potential energy of hanging mass

= M1*g*d

= 150*9.8*5

= 7350 J

This Energy is used to increase the temperature.

so, W = m*c*dT (here m is mass of water, c is specific heat of water and dT is the increase in temperature)

dT = W/(m*c)

= 7350/(5*4186)

= 0.35 degrees celsius

b) Workdone on water = decrease in potential energy of M1 and M2

= (M1+M2)*g*d

= (150+150)*9.8*5

= 14700 J

This Energy is used to increase the temperature.

so, W = m*c*dT (here m is mass of water, c is specific heat of water and dT is the increase in temperature)

dT = W/(m*c)

= 14700/(5*4186)

= 0.7 degrees celsius

c) Workdone on water = decrease in potential energy of M1 and M2

= (M1+M2)*g*d

= (75+150)*9.8*5

= 11025 J

This Energy is used to increase the temperature.

so, W = m*c*dT (here m is mass of water, c is specific heat of water and dT is the increase in temperature)

dT = W/(m*c)

= 11025/(5*4186)

= 0.53 degrees celsius

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