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EastWind [94]
3 years ago
15

In a calorimetry experiment, three samples A, B, and C with TA> TB> Tc are placed in thermal contact. When the samples hav

e reached thermal equilibrium at a common temperature T, which one of the following must be true?
a. QA > QB >QC
b. QA< 0, QB <0, and Qc > 0
c. T> TB
d. T e. TA > T> Tc
Physics
1 answer:
zaharov [31]3 years ago
3 0

Answer:

e. TA>T>Tc

Explanation:

a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:

Q=mC_{p}(T_{f}-T_{0})

if we did an energy balance of the system, we would get that>

QA+QB+QC=0

For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.

We don't know either of them, so we cannot determine if this statement is true.

b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.

c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.

d) is not complete

e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.

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A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
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Answer:

20J

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In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

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Where;

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From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

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The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

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