Answered: m2, T, T2 m. T; =f m2 m3

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter1: Introduction To Statics

Section: Chapter Questions

Problem 1.31P: Resolve the 360-lb force into components along the cables AB and AC. Use =60 and =40.

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Question

The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form where m3 and T3 are the mass and temperature of the final mixture, respectively

Expert Solution

Step 1

Given:

The mass of the first gas is m₁.
The mass of the second gas is m₂.
The temperature of both the gases is T₁ and T₂.
The mass and temperature of the mixture of the gases are m₃ and T₃.

The relation between the mass of mixture and gases is,

$m_{3} = m_{1} + m_{2}$

The expression for the energy equation is,

$E_{i n} + E_{o u t} = ∆ E_{s y s t e m}$

Here E_in and E_out are the energies inside and outside the system and ∆E_system is the change in energy of the system.

Since it’s in closed system energy transferred outside the system is zero and then change in internal energy will be zero.

$\begin{array}{rcl} ∆ U & = & 0 \\ m_{1} C_{V} (T_{3} - T_{1}) + m_{2} C_{v} (T_{3} - T_{2}) & = & 0 \\ m_{1} C_{V} T_{1} + m_{2} C_{v} T_{2} & = & (m_{1} + m_{2}) C_{v} T_{3} \end{array}$

Here ∆U is the change in internal energy.

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