Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 2

Q. 2.25

A system consists of a stone of mass 10 kg and a bucket containing 220 kg of water having same temperature. The stone falls into water from a height of 100 m. Calculate (a) the change in internal energy, (b) change in kinetic energy, (c) change in potential energy, (d) heat transferred, and (e) work transferred for the following cases:

(i) The instant the stone is about to enter water.
(ii) Just after the stone comes to rest in the bucket.
(iii) The temperature of stone and water remains the same.

Step-by-Step

Verified Solution

(i) According to the first law of thermodynamics,

δQ – δW = dU + d(KE) + d(PE)

If δQ = 0 and δW = 0, then

dU + d(KE) + d(PE) = 0

For a non-flow process, dU = 0

\therefore \quad d(KE) = – d(PE)

Or \quad \Delta K E=-\Delta P E=-m g\left(z_2-z_1\right)

= – 10 × 9.81(0 – 100) = 9810 J or 9.81 kJ

ΔPE = – 9.81 kJ

(ii)               δQ = 0, δW = 0 and d(KE) = 0

 d(PE) = – dU

Or               ΔPE = – ΔU

\Delta P E=m g\left(z_2-z_1\right)=10 \times 9.81(0-100)=-9.81 \ kJ

(iii)               dU = 0, d(KE) = 0, and  δW = 0

\therefore \quad      Q_{1-2}=\Delta P E=m g\left( z _2- z _1\right)=-9.81 \ kJ