Answer:
The value is [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]
Explanation:
From the question we are told that
The no of moles of the sample is n = 0.0447 mole
The formula weight is [tex]M = 114 \ g/mol[/tex]
The mass of water is [tex]m = 6.19 *10^{2}\ g[/tex]
The amount of the fuel is [tex]f= 6.13*10^{-1} \ nutritional \ Cal[/tex]
The temperature rise is [tex]\Delta T = 5.05^o[/tex]
Generally
[tex]1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg[/tex]
=> [tex]f= 6.13*10^{-1} \ nutritional \ Cal \to x[/tex]
=> [tex]x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}[/tex]
=> [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]
