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Furkat [3]
3 years ago
5

If 190 grams of water is cooled from 42.7°C to 21.2° C how much energy was lost by the water?

Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

Energy lost by the water is 17.1 x 10³ J .

Explanation:

Specific heat is defined as the amount of energy per unit mass needed to raise the temperature of the substance by a one degree Celsius.

Heat energy gain or loss by any substance is given by :

Q = m x C x ( T₁ - T₂ )     .....(1)

Here m is the mass of the substance, C is specific heat of the substance and T₁ and T₂ are the initial and final temperature of the substance.

According to the problem,

Mass of water, m = 190 gm

Specific heat of water, C = 4.186 J gm⁻¹ ⁰C⁻¹

Initial temperature, T₁ = 42.7⁰ C

Final temperature, T₂ = 21.2⁰ C  

Substitute these values in equation (1).

Heat energy loss by water = 190 x 4.186 x ( 21.2 - 42.7 )

                                            =  -17.1 x 10³ J

In the above value, negative sign denotes the loss in energy.

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A total of 875 J of work was done when a force of 175 N was exerted on a box to move it. How far was the box moved
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2 years ago
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
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Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
The second minimum in the diffraction pattern of a 0.11-mm-wide slit occurs at 0.72°. What is the wavelength of the light?
Anit [1.1K]

Answer:

691.13 nm

Explanation:

d = width of the slit = 0.11 x 10⁻³ m

θ = angle of diffraction pattern = 0.72° degree

λ = wavelength of the light = ?

m = order = 2                              (since second minimum)

for the second minimum diffraction pattern we use the equation

d Sinθ = m λ

Inserting the values

(0.11 x 10⁻³) Sin0.72 = (2) λ

λ = 691.13 x 10⁻⁹ m

λ = 691.13 nm

6 0
3 years ago
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