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Nitella [24]
3 years ago
10

How much thermal energy is needed to raise the temperature of 15kg gold from 45⁰ C up to 80⁰ C​

Engineering
1 answer:
Levart [38]3 years ago
6 0

Answer:

68.25 kJ

Explanation:

The thermal energy Q required to raise the temperature of 15kg gold from 45⁰ C up to 80⁰ C​ is Q = mcΔθ where m = mass of gold = 15 kg, c = specific heat capacity of gold = 130 J/kg°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = 45 °C and θ₂ = 80 °C

So, Q = mcΔθ = mc(θ₂ - θ₁)

= 15 kg × 130 J/kg°C × (80 °C - 45 °C)

=  1950 J/°C × 35 °C

= 68,250 J

= 68.25 kJ

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Explanation:

Concentration overpotential, ηc,

I hope it helps you

5 0
3 years ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
3 years ago
A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition
zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

V_{f}=0.61

V_{m}=1-V_{f}

     =1-0.61

     =0.39

E_{f}=10 \ Mpa

\sigma_{f}=0.35 \ Mpa

E_{m}=0.45 \ Mpa , \sigma_{m}=9\times 10^{-3} \ Mpa

As we know,

⇒  E_{e}=fE_{f}+mE_{m}

On putting the estimated values, we get

⇒       =0.61\times 10+0.39\times 0.95

⇒       =6.27 \ Mpa

Now,

⇒  \sigma_{c}=f\sigma_{f}+m\sigma_{m}

On putting the estimated values, we get

⇒       =0.61\times 0.35+0.39\times 0.009

⇒       =0.217 \ Mpa

Therefore,

The load carried by fiber,

=\frac{f\sigma_{f}}{\sigma_{c}}

=\frac{0.35\times 0.61}{0.217}

=0.98 i.e., 98%

4 0
4 years ago
Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the
geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

a)

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats  because it is less dense than water

b)

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats   because it is less dense than water

c)

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks   because it is  denser than water

4 0
3 years ago
A zener diode exhibits a constant voltage of 5.6 volts for currents greater than five times the knee current.Izk is specified to
iren2701 [21]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

Download docx
5 0
3 years ago
Read 2 more answers
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