1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anastasy [175]
3 years ago
6

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

ater and its container are heated to 70oC. Assuming no evaporation, what then will be the depth of the water column if the coefficient of thermal expansion for the glass is 3.8*10-6 mm/mm peroC ?
Engineering
1 answer:
Dafna11 [192]3 years ago
4 0

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

You might be interested in
Consider 2 hosts, Host A and Host B, transmitting a large file to a Server C over a bottleneck link with a rate of R kbps.
nevsk [136]

Answer:

i want coins sorry use a calculator or sum

Explanation:

kk

5 0
3 years ago
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th
ddd [48]

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

4 0
3 years ago
In a website browser address bar, what does “www” stand for?
Ludmilka [50]

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

3 0
3 years ago
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Levart [38]

Answer:

As the asteroid falls closer to the Earth's surface its <u>Gravitational</u> <u>Potential</u> energy <em>decreases</em> and its <u>Kinetic</u> energy <em>increases</em>.

4 0
3 years ago
A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow
kotegsom [21]

Answer:

(d) None. No provisions exist.

Explanation:

B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.

5 0
3 years ago
Other questions:
  • In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab
    14·1 answer
  • What would happen to a plane if the weight force becomes greater than the lift force?
    12·2 answers
  • Disc brake rotors that are too thin cannot handle as much heat and will experience ___________.
    6·1 answer
  • Who plays a role in the financial activities of a company?
    10·1 answer
  • A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
    15·1 answer
  • What is 94*738^389428394
    8·1 answer
  • Compared with space operations specialists, intelligence officers are which of the following?
    7·1 answer
  • How wold you classify the earliest examples of S.T.E.M discoveries provided in this lesson?
    11·1 answer
  • Problema sobre programacion orientada a objetos!!
    14·1 answer
  • Write down the tracking error such that the adaptive cruise control objective is satisfied.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!