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]
=> [tex]Q = (0.5 * 4186 + 0.1 * 900) * 54.9[/tex]
=> [tex]Q = 119847 \ J[/tex]
Generally the energy generated per kilogram of dried algae is mathematically represented as
[tex]E = \frac{Q}{m}[/tex]
=> [tex]E = \frac{119847}{0.005}[/tex]
=> [tex]E = 2.396 *10^{7} \ J/kg[/tex]
