A piece of metal has a mass of 10g and a mass of 2cm

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

Harrizon [31]

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

4 0

2 years ago

Which is an organic compound

DedPeter [7]

Organic compound, any of a large class of chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides.

3 0

2 years ago

How many neutrons does an element have if its atomic number is 45 and its mass number is 161

disa [49]

Here is the formula that you need to know for this problem:
Number of proton (atomic number) + Number of neutrons = Mass number

So,

45 + Number of Neutrons = 161

Number of neutrons = 161 - 45

Number of Neutrons = 116

5 0

3 years ago

Read 2 more answers

If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant for

densk [106]

Answer:

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

Part B

The buoyant force is approximately 40.93 N

Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg

The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.

7 0

3 years ago

2)Two masses M move at speed V, one to the east and one to the west. What is the total energy of the system?a.Now consider the s

sp2606 [1]

Answer:

Part A: E = MV²

Part B: E1/3 = energy of the first car relative to the moving frame = 1/2×M×(V + u)²,

E2/3 = energy of the first car relative to the moving frame= 1/2×M×(u – V)²

Part C: Et/3 = Total Energy of the System = M(V² + u²)

Explanation:

Part A

Taking the east direction as positive,

Vehicle 1: mass = M, velocity = V

Vehicle 2: mass = M, velocity = – V (west)

In the first case where both vehicles are viewed from a stationary frame, the total energy of the system is

E = 1/2×MV² + 1/2×M(–V)²

E = 1/2×MV² + 1/2×MV²

E = MV²

Part B

Now Case 2: moving frame

Vehicle 1: mass = M, velocity = V

Vehicle 2: mass = M, velocity = – V (west)

Vehicle 3: velocity = – u (west)

For this case we need to find the velocities of vehicles 1 and 2 relative to vehicle 3

Let V1/3 be the velocity of vehicle 1 relative to 3 and

V2/3 be the velocity of vehicle 2 relative to 3

V1/e = velocity of vehicle 1 relative to the earth = V

V2/e = velocity of vehicle 2 relative to the earth = – V

V3/e = velocity of vehicle 2 relative to the earth = – u

(still choosing the east direction as positive)

Ve/3 = velocity of earth relative to vehicle 3 = – V3/e = – (–u) = u

So taking the earth as a stationary frame of reference,

V1/3 = V1/e + Ve/3

V1/3 = V + u

V2/3 = V2/e + Ve/3

V2/3 = –V + u

V2/3 = u – V

E1/3 = 1/2×M×V1/3²

E1/3 = 1/2×M×(V + u)²

E1/3 = 1/2×M×V1/3²

E2/3 = 1/2×M×(u – V)²

Part C

Let

Et/3 = Total energy relative to the moving frame = 1/2×M×(V + u)² + 1/2×M×(u – V)²

Et/3 = 1/2×M×(V² + 2Vu +u² + u² – 2Vu + V²)

Et/3 = 1/2×M×(2V² + 2u²)

Et/3 = 1/2×2×M×(V² +u²)

Et/3 = M(V² + u²)

The total energy of the system increases by Mu²

6 0

3 years ago