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
arlik [135]
3 years ago
12

330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

s 840 grams and initial temperature 29°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water?
Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

You might be interested in
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
3 years ago
Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt
Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: 800* 24 = 1,92 *10^4kg

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

4 0
2 years ago
Betelgeuse is how many times larger than the sun
Aleksandr-060686 [28]
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
3 0
3 years ago
The ___ energy in a mechanical system is determined by adding the potential and kinetic enters together
Svetllana [295]

Answer:

A, total.

<em>The </em><em>total</em><em> energy in a mechanical system is determined by adding the potential and kinetic enters together.</em>

<em />

<u><em>i hope this helped at all.</em></u>

<em />

4 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
Other questions:
  • Answer fast please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
    15·2 answers
  • A forensic scientist receives an unknown liquid. Upon close observation, it appears there may be small objects floating in the l
    9·1 answer
  • Plasma is a controllable reactive gas that is used to make small ____ in silica which are used in computers and cell phones.
    15·1 answer
  • Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolutio
    13·2 answers
  • A car starts off at a 0mph the final velocity is 203mph and the acceleration is an AVERAGE of 2.93ft/s^2 and it is traveling up
    9·1 answer
  • Which of the following is a chemical property of matter?
    11·2 answers
  • What will be the average velocity of a body falling in free fall on Earth for 3 s?
    7·1 answer
  • Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t
    6·1 answer
  • What is a question that an engineer might have about a can opener?
    15·1 answer
  • hypnotist covers a distance of 5 kilometre in one hour and Rahul covers the same distance in 2 hour calculate the speed and stat
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!