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Darya [45]
2 years ago
14

A student has a beaker containing 55g of water at 100°c. how much heat is needed to convert the water to steam

Physics
2 answers:
masha68 [24]2 years ago
5 0
Answer is: 124135 J.
m(water) = 55g.
T(water) = 100°C.
Q = ?
Q - heat needed to convert water to steam.
∆Hvap = 2257 J/g
∆Hvap - enthalpy of vaporization is the amount of energy <span> needed to transform the liquid substance into a gas.
Q = </span>∆Hvap · m(water) = 2257 J/g · 55g = 124135 J.

tamaranim1 [39]2 years ago
4 0
In this case, the water is at 100Degrees and standard pressure so we are already at the boiling point.
 So we only have to do the change phase.
 Heat of vaporization for water (GOOGLE) 2260 J/g.
 So, 55*2260=124300 J.
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Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
djverab [1.8K]
Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do.  (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)

You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use =   (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone.  It also helps you appreciate
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burn for 105 hours (more than 4-1/2 days ! ! !)

5 0
2 years ago
Read 2 more answers
A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
  2. Particle's speed and direction of motion brainly.com/question/2814900
  3. About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
2 years ago
Read 2 more answers
A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati
fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0
2 years ago
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

\frac{1}{f} = 0.2(\frac{1}{30})

f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

5 0
2 years ago
Read 2 more answers
2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and
Georgia [21]

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

5 0
2 years ago
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