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Svetach [21]
2 years ago
15

Two equal quantities of water, of mass m and at temperatures T1 and T2, (T1 > T2) are mixed together with the pressure kept c

onstant. Assume heat exchange occurs only between the two water containers. (a) Show that the entropy change of the universe is delta S = 2mc_P ln(T_1 + T_2/2 squareroot T_1T_2), where c_P is the specific heat capacity of water at constant pressure. Show that delta S > 0 for any finite temperatures T_1 and T_2.

Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))

Explanation:

The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.

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Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

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We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

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a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

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3 years ago
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