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3 years ago

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A 75-kg copper block initially at 110°C is dropped into an insulated tank that contains 160 L of water at 15°C. Determine the fi

nal equilibrium temperature and the total entropy change for this proces

1 answer:

joja [24]3 years ago

3 0

Answer:

T =43.61⁰C

ΔS = 0.84 KJ/K

Explanation:

Given that

Mass of copper block m₁ = 75 kg

Initial temperature of copper block ,T₁ = 110⁰C

Initial temperature of copper block ,water T₂ = 15⁰C

Specific heat for copper ,Cp₁=0.385 J/KkgK

Specific heat for water,Cp₂=4.187 KJ/kgK

Mass pf water ,m₂ = 0.16 x 1000 = 16 kg

Let's take final temperature = T

Heat lost by copper block = Heat gain by water

m₁ Cp₁ (T₁ - T) = m₂Cp₂(T-T₂)

$T=\dfrac{m_1C_{P1}T_1+m_2C_{P2}T_2}{m_1C_{P1}+m_2C_{P2}}$

$T=\dfrac{75\times 0.385\times (273+110)+16\times 4.187\times (273+15)}{75\times 0.385+16\times 4.187}$

T=316.61 K

T =43.61⁰C

The total change in entropy

$\Delta S=m_1C_{P1}\ln\dfrac{T}{T_1}+m_2C_{P2}\ln\dfrac{T}{T_2}$

$\Delta S=75\times 0.385\times\ln\dfrac{316.61}{273+110}+16\times 4.187\times \ln\dfrac{316.61}{273+15}$

ΔS = 0.84 KJ/K

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3 years ago

Two trains of length 90m and120m<br>

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Answer:Train A is 90 m long and running at 72 km/h.

Train B is 120 m long and running at 36 km/h.

When the trains are running in opposite directions, their relative speed gets added: 72+36 = 108 km/h.

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4 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

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Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

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If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

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Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

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Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

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