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Alexus [3.1K]
3 years ago
6

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

ance A is 6.01 g and its initial temperature is 20.0 degrees Celsius. The mass of substance B is 25.6 g and its initial temperature is 52.2 degrees Celsius. The final temperature of both substances at thermal equilibrium is 46.1 degrees Celsius. If the specific heat capacity of substance B is 1.17 J/g degrees Celsius, what is the specific heat capacity of substance A
Engineering
1 answer:
Kaylis [27]3 years ago
5 0

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

<u>FOR SUBSTANCE A:</u>

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

<u>FOR SUBSTANCE B:</u>

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

<u>C = 1.16 J/g</u>

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Answer:

The answer is "3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}"

Explanation:

Given data:  

Initial temperature of tank T_1 = 300^{\circ}\ C= 573 K

Initial pressure of tank P_1= 400 \ kPa

Diameter of throat d* = 2 \ cm

Mach number at exit M = 2.8

In point a:

calculating the throat area:

A*=\frac{\pi}{4} \times d^2

      =\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2

Since, the Mach number at throat is approximately half the Mach number at exit.  

Calculate the Mach number at throat.  

M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4

Calculate the exit area using isentropic flow equation.

\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}}  (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}

Here: \gamma is the specific heat ratio. Substitute the values in above equation.

\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}}  (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K

Calculate the velocity at exit.  

V*=M*\sqrt{ \gamma R T*}

Here: R is the gas constant.  

V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}

Calculate the density of air at inlet

\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\  \frac{kg}{m^3}

Calculate the density of air at throat using isentropic flow equation.  

\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}

Calculate the mass flow rate.  

m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}

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<u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u>

Explanation:

<h2>Given:</h2>

Acceptable volume = 40,000 cm³

Area of luggage = 1,960 cm²

Height of luggage = 23 cm

<h2>Question:</h2>

Is the passenger's luggeage ok to carry onto the airplane

<h2>Equation:</h2>

V = l x w x h

or we can use

V = A x h

Since A = l x w

where: V - volume

A - area

l - length

w - width

h - height

<h2>Solution:</h2>

V = A x h

V = ( 1,960 cm²)(23 cm)

V = 45,080 cm³

45,080 cm³ is greater than the acceptable volume 40,000 cm³

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