If you and your friend EACH apply 55N of force pushing against a wall for 10 minutes. How much work

In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g

Veseljchak [2.6K]

Answer:

The mass of the ice added = 16.71 g

Explanation:

The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

Heat gained by the ice = Heat lost by the 326 g of water.

Let the mass of ice be m

The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

m = unknown mass of ice

C = Specific Heat capacity of ice = 2.108 J/g°C

ΔT = change in temperature = 0 - (-5.5) = 5.5°C

Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

Heat used by the ice to melt at 0°C = mL

m = unknown mass of ice

L = Latent Heat of fusion of ice to water = 334 J/g

Heat used by the ice to melt at 0°C = m×334 = (334m) J

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

m = unknown mass of water (which was ice)

C = Specific Heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 15 - 0 = 15°C

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

Heat lost by the water in the calorimeter cup = MCΔT

M = mass of water in the calorimeter cup = 326 g

C = specific heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 20 - 15 = 5°C

Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

Heat gained by the ice = Heat lost by the 326 g of water.

408.384m = 6,823.18

m = (6,823.18/408.384)

m = 16.71 g

Hope this Helps!!!

7 0

3 years ago

A crane lifts 1800kg mass through a vertical height of 6cm in 9 seconds, Taking (g) as 10N/kg. what is the cranes power output?

valkas [14]

Explanation:

power output=(1800×10×0.06)/9=120watts

6 0

3 years ago

An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy a

Shtirlitz [24]

Explanation:

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

$\bold{↬{ }}$ ${\bold{\sf{\underline{Given \: that}}}}$

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

$\bold{↬{ }}$ ${\bold{\sf{\underline{To \: find}}}}$

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

${\bold{\sf{\underline{Solution}}}}$

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

${\bold{\sf{\underline{Using \: concepts}}}}$

✰ Potential energy formula.

${\bold{\sf{\underline{Using \: formula}}}}$

✰ Potential energy = mgh

${\bold{\sf{\underline{We \: also \: write \: these \: as}}}}$

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

${\bold{\sf{\underline{Full \: solution}}}}$

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0

3 years ago

Read 2 more answers

Describe succinctly the relationship between how far a galaxy is from us (its distance), versus how fast it is moving.

Eva8 [605]

Answer:

Distance is directly proportional to the velocity

Explanation:

In 1929, Edwin Hubble's wrote an article that talked about relationship between the distance and recession speed/velocity of galaxies which led to what is known as the Hubble Law. This law states that galaxies are moving away from the earth at velocities proportional to their distances.

Thus is written as;

v = H_o•d

Where;

v is velocity

d is distance

H_o is Hubble's constant rate of cosmic expansion.

He came to this conclusion by generating a graph known as Hubble's classic graph which was a graph of observed velocity vs distance for nearby galaxies.

7 0

3 years ago

For a reaction to occur what must happen to the energy in order to break the chemical bond

Andrej [43]

Energy needs to realease

6 0

3 years ago