Question

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 mainland.
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?

Answer 1

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

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