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kirill [66]
3 years ago
9

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Physics
1 answer:
Effectus [21]3 years ago
5 0

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

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discrimination: prejudice towards a person/group based on their race, sex, age, and/or sexual orientation

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People of color face discrimination because of the color of their skin.

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If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta
Paraphin [41]

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

d=v_ot+\dfrac{1}{2}at^2

Since the initial velocity of a dropped object is 0, we can make this the equation:

d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900

t=30\text{ seconds}

The final velocity can be calculated with the formula:

v_f=v_o+at

Once again, since there is no initial velocity:

v_f=at \\\\v_f=(9.8)(30)=294m/s

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

5 0
3 years ago
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
4 years ago
Suppose you now take the ball and using a bat, pop it straight up into the air with a hang-time of 5.00 s (the hang time is how
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Answer:

<h3>30.66m</h3>

Explanation:

Using the equation of motion formula S = ut + \frac{1}{2}gt^2 where;

S is the height to which the ball rises

u is the initial velocity of the ball = 0m/s

a is the acceleration due to gravity = 9.81m/s²

t is the time taken by the ball in air = 5.0s

Note that the  time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s

Substituting the given parameters into the formula above to get S:

S = ut + \frac{1}{2}gt^2\\\\S = 0(2.5)+ \frac{1}{2}(9.81)(2.5)^2\\\\S = 0+\frac{1}{2}(9.81)\times 6.25 \\\\S = \frac{61.3125}{2}\\ \\S = 30.65625m\\\\S \approx 30.66m

This means that the ball rises 30.66m before it reaches its peak.

8 0
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