Answer:
The  value is  [tex]E =  2.396 *10^{7} \  J/kg[/tex]
Explanation:
From the question we are told that
The specific heat of water is  [tex]c_w =  4186 \  J/(kg \cdot K)[/tex]
The specific heat of  aluminum is  [tex]c_a  =  900 \  J/(kg \cdot K)[/tex]
 The  temperature increase is  [tex]\Delta  T  =  54.9^oC[/tex]
 The  mass of water  is  [tex]m_w =  0.500 \  kg[/tex]
The  mass of the aluminum is  [tex]m_a  =  0.100 \ kg[/tex]
The  mass of the algae pellet is  [tex]m  =  5.00 \  g  =  0.005 \  kg[/tex]
The  heat transfer is mathematically represented as
   [tex]Q =  (m_w * c_w + m_a *  c_a) \Delta  T[/tex]
=> Â [tex]Q = Â (0.5 * 4186 + Â 0.1 * 900) * Â 54.9[/tex]
=> Â [tex]Q = Â 119847 \ Â J[/tex]
Generally the energy generated per kilogram of  dried algae is mathematically represented as
    [tex]E = \frac{Q}{m}[/tex]
=> Â Â [tex]E = Â \frac{119847}{0.005}[/tex]
=> Â Â [tex]E = Â 2.396 *10^{7} \ Â J/kg[/tex]
