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Oduvanchick [21]

3 years ago

10

An 8.5 g ice cube is placed into 255 g of water. Calculate the temperature change in the water upon the complete melting of the

ice. Assume that all of the energy required to melt the ice comes from the water.

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

Change in temperature of water = 2.67 °C

Explanation:

Heat gained by ice = heat lost by water

Q₁ = Q₂............... Equation 1

Where Q₁ = Latent heat of ice, Q₂ = heat lost by water

Q₁ = lm₁.................... equation 2

Q₂ = cm₂ΔT ............... Equation 3

Substituting equation 2 and 3 into equation 1

lm₁ = cm₂ΔT............... Equation 4

Making ΔT the subject of formula in equation 4

ΔT = lm₁/cm₂............. Equation 5

Where ΔT = change in temperature of water, l = specific latent heat of ice, m₁ = mass of ice, c= specific heat capacity of water, m₂ = mass of water.

Given: m₁ = 8.5 g = 0.0085 kg, m₂ = 255 g = 0.255 kg.

Constant: l = 336000 J/kg, c = 4200 J/kg.K

Substituting these values into equation 5

ΔT = (336000×0.0085)/(0.255×4200)

ΔT = 2856/1071

ΔT = 2.67 °C

Change in temperature of water = 2.67 °C

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Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

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- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

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In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

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3 years ago

I don’t understand the 4th one please help someone.

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Answer:

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BC=8

CD=4

DE=8

EF=4

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Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the

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$W_b = F_b d$

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Answer:

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Putting the values

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4 years ago