All categories

2 years ago

How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams

Physics

1 answer:

Harrizon [31]2 years ago

4 0

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

H = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

H = 20g x 2260J/g

H = 45200J

You might be interested in

Give at least one fact about subatomic particles

victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

- These particles can be divided as Brayons and Leptons

- These particles are often held together by one of the four fundamental particles ( Weak force, strong force, electromagnetic force, gravitational force).

6 0

2 years ago

1. Since sleep is so important, we might wonder why people so often fail to get a sufficient amount

ziro4ka [17]

Answer: 9/10

Explanation:

because it's really important and makes you energetic

8 0

2 years ago

The force of air particles over an area is A.) temperature B.) pressure C.) Volume D.) Kelvins

Westkost [7]

Answer:The anwser is B i Promise ok

Explanation:

3 0

2 years ago

Read 2 more answers

Explain how evaporation cools a liquid

MAVERICK [17]

When evaporation occurs liquid absorbs heat from the surroundings to get converted to its vapour form as a result, there is an overall decrease in the heat leading to cooling of the liquid.

Hope that this was helpful :)

8 0

2 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago