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Komok [63]
4 years ago
6

How much heat is required to warm 1.40 l of water from 20.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.)?

Physics
1 answer:
marta [7]4 years ago
3 0
The water density is 
d=1.0 g/mL = 1000 g/L
And the mass of 1.40 L of water is
m=dV=(1000 g/L)(1.40 L)=1400 g

The amount of heat needed to increase the temperature of the water by \Delta T is given by
Q=m C_s \Delta T
where m is the water mass, C_s = 4.18 J/g ^{\circ}C is the water specific heat capacity and 
\Delta T=100.0 ^{\circ}C-20.0 ^{\circ}C = 80.0^{\circ}C 
is the increase in temperature. If we substitute these numbers into the equation, we find
Q=(1400 g)(4.18 J/g^{\circ}C)(80.0^{\circ}C)=4.68 \cdot 10^5 J
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A block of inertia m is placed on an inclined plane that makes an angle θ with the horizontal. The block is given a shove direct
Wittaler [7]

Answer:

A) d = v² / (2g (μ cos θ + syn θ)     B)    μ = tan θ

Explanation:

Part A

We can work this part with the work and energy theorem, where the work of the friction forces is equal to the energy change of the system.

The work is

       W = fr .d

With the force of friction it opposes the movement

       W = - fr d

The energy at the lowest point is

      Em₀ = K = ½ m v²

The energy at the highest point

      Em_{f} = U = m g y

The height (y) can be found by trigonometry

      sin θ = y / d

      y = d sin θ

     W =  Em_{f} –Em₀

     -fr d = mg d sin θ - ½ m v²

The equation for the force of friction is

      fr = μ N

From Newton's second law

      N - W cos Te = 0

We replace

     -μ (mg cos θ) d - mg d sin θ = - ½ m v²

      d g (μ cos θ + sin θ) = ½ v²

      d = v² / (2g (μ cos θ + syn θ)

Part B

The block is stopped, what is the Angle tet, let's use Newton's second law

      fr - W sin θ = 0       ⇒     fr = W sin θ

      N - W cos θ= 0       ⇒    N = w cos θ

      fr = μ N

      μ (mg cos θ) = mg syn θ

      μ = syn θ / cos θ

       μ = tan θ

7 0
3 years ago
If a substance besides water is changed from a gas to a liquid and then to a solid, what happens to the particles?
NARA [144]
The correct answer is they get closer together

Going from gas into a liquid makes the molecules move closer together, which happens again when you turn it into a solid.
3 0
3 years ago
Read 2 more answers
An object is 27.0 cm from a concave mirror of focal length 15.0 cm. find the image distance.
pantera1 [17]

The distance of the Image will be -33.75 cm

A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.

Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm

We have to find distance of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-15) = 1/v + 1/(-27)

1/(-15) = 1/v - (1/27)

1/v = -0.029

v = -33.75 cm

Therefore the distance of the Image will be -33.75 cm

Learn more about concave mirror here:

brainly.com/question/9816370

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8 0
2 years ago
A 100 kilowatt bulb burns for 3 hours how much energy did it use
belka [17]

Answer:

1.08×10⁹ J

Explanation:

From the question given above, the following data were obtained:

Power (P) = 100 kilowatt

Time (t) = 3 hours

Energy (E) =?

Next, we shall convert 100 kilowatt to Watts. This can be obtained as follow:

1 KW = 1000 W

Therefore,

100 KW = 100 KW × 1000 W / 1 KW

100 KW = 100000 W

Thus, 100 KW is equivalent to 100000 W.

Next, we shall convert 3 hrs to second (s). This can be obtained as follow:

1 h = 3600 s

Therefore,

3 h = 3 h × 3600 s / 1 h

3 h = 10800 s

Thus, 3 h is equivalent to 10800 s.

Finally, we shall determine the amount of energy consumed as follow:

Power (P) = 100000 W

Time (t) = 10800 s

Energy (E) =?

P = E/t

100000 = E / 10800

Cross multiply

E = 100000 × 10800

E = 1.08×10⁹ J

Therefore, 1.08×10⁹ J of energy was consumed.

4 0
3 years ago
Calculate the energy needed to change 200.0 g of ice from -20.0°C to water at 35.0°C.
oksano4ka [1.4K]

it's C  24.94KCAL

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3 0
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