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3 years ago

8

In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainl

and.
Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?

1 answer:

ozzi3 years ago

7 0

Answer:

a

$Q = 5.34 *10^{19} \ J$

b

$T = 0.445 * 365 = 162. 413 \ days$

Explanation:

From the question we are told that

The area of Manhattan is $a_k = 87.46 *10^{6} \ m^2$

The area of the ice is $a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2$

The thickness is $t = 500 \ m \\$

Generally the volume of the ice is mathematically represented is

$V = a_i * t$

substituting value

$V = 500 * 3.498*10^{8}$

$V = 1.75 *10^{11}\ m^3$

Generally the mass of the ice is

$m_i = \rho_i * V$

Here $\rho_i$ is the density of ice the value is $\rho _i = 916.7 \ kg/m^3$

=> $m_i = 916.7 * 1.75*10^{11}$

=> $m_i = 1.60 *10^{14} \ kg$

Generally the energy needed for the ice to melt is mathematically represented as

$Q = m _i * H_f$

Where $H_f$ is the latent heat of fusion of ice and the value is $H_f = 3.33*10^{5} \ J/kg$

=> $Q = 1.60 *10^{14} * 3.33*10^{5}$

=> $Q = 5.34 *10^{19} \ J$

Considering part b

We are told that the annual energy consumption is $G = 1.2*10^{20 } \ J / year$

So the time taken to melt the ice is

$T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}$

$T = 0.445 \ years$

converting to days

$T = 0.445 * 365 = 162. 413 \ days$

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Answer:

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Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

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Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

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$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

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Explanation:

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3) and 4)

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