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
AleksAgata [21]
3 years ago
14

1.80 kJ of heat is added to a slug of gold and a separate 1.80 kJ of heat is added to a slug of manganese. The heat capacity of

the gold slug is 149J/°C while the heat capacity of the manganese slug is 505 J/°C. If the slugs are each originally at 25.00°C, what is the final temperature of each slug?
Physics
1 answer:
Svetlanka [38]3 years ago
7 0

Answer:

final temperature of  slug of gold is 37°C

final temperature of  slug of manganese is 28.56°C

Explanation:

Hello! To solve this problem we must take into account the concept of heat capacity, this is defined as the ratio between energy and temperature rise.

In other words it is the amount of energy that is required to increase a temperature degree.

Taking into account the above we infer the following equation

where  

C=heat capacity

You might be interested in
An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t
SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f)  in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0
3 years ago
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
mrs_skeptik [129]

Answer:

25 to the right

Explanation:

there you go friend your awsome

4 0
2 years ago
An important aspect of fission reactions is that they produce _______________ which allow for chain reactions.
Alex777 [14]

An important aspect of fission reactions is that they produce free neutrons,  which causes chain reactions.

4 0
2 years ago
Read 2 more answers
George and Harriot walk with an average velocity of .95 m/s eastward. If it takes them 30
artcher [175]

Answer:

1.71 km

Explanation:

Convert 30 minutes to seconds:

30 min × (60 s / min) = 1800 s

Find the displacement:

0.95 m/s × 1800 s = 1710 m

Convert to kilometers:

1710 m × (1 km / 1000 m) = 1.71 km

5 0
3 years ago
Read 2 more answers
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi
elena55 [62]
The angular acceleration of a rotating object is given by
\alpha =  \frac{\omega_f - \omega_i}{\Delta t}
where
\omega_f is the final angular speed of the object
\omega_i is its initial angular speed
\Delta t is the time taken to accelerate

For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2
8 0
2 years ago
Other questions:
  • To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres
    9·1 answer
  • table has several directional compasses, several lengths of wire, an iron nail, a battery, an ammeter, a light bulb, a permanent
    12·1 answer
  • A 48-kg woman pushes a 12-kg grocery cart with a force of 24 N. What is the magnitude of the force that the grocery cart exerts
    12·1 answer
  • I need help wit this physics question.
    15·1 answer
  • The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part
    11·2 answers
  • How many meters across is a road sign that has an angular size of 120 arcseconds and is 1 km away?
    5·1 answer
  • How do mathematical models help us learn about conditions inside the sun?
    5·1 answer
  • Explain why position is a dependent variable in a position versus time graph.
    11·1 answer
  • What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87 m?​
    12·1 answer
  • Two pulses move in opposite directions on a string and are identical in shape except that one has positive displacements of the
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!