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Art [367]
2 years ago
13

WhAT is the change IN THE entropy of 2.0kg of h2o molecules when transform at a constant pressure of 1 atm from water at 100 deg

rees celcius to steam at the same tEMPERATURE?
Physics
1 answer:
Fynjy0 [20]2 years ago
3 0

Answer:

The entropy change is 45.2 kJ/K.

Explanation:

mass of water at 100 C = 2 kg

Latent heat of vaporization, L = 2260 kJ/kg

Heat is

H = m L

H = 2 x 2260 = 4520 kJ

Entropy is given by

S = H/T = 4520/100 = 45.2 kJ/K

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It was once recorded that a Jaguar
Artyom0805 [142]

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

6 0
3 years ago
What type of epithelial tissues is in the liver
snow_lady [41]

The glandular epithelial tissues make up the liver.

4 0
3 years ago
Read 2 more answers
Calculate the power of convex lens of focal length 20cm​
maxonik [38]

D = 1/f, where D is the power in diopters and f is the focal length in meters.

D=1/20

<u>D=0.05</u>

6 0
3 years ago
Can u help me with C
Vladimir79 [104]
I believe it would be 2m/s.
6 0
3 years ago
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A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 years ago
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