Answer:
[tex]200\frac{kcal}{g}[/tex]
Explanation:
Hello!
In this case, for this calorimetry problem, since the combustion 0.0500 mol of the nutrient increase the temperature of water by 5.70 °C, we can notice that the heat lost by the nutrient is gained by water in order to write:
[tex]Q_{nutrient}=-Q_{water}[/tex]
Which can be also written as:
[tex]Q_{nutrient}=-m_{water}C_{water}\Delta T_{water}\\\\Q_{nutrient}=-200 g*4.184\frac{J}{g\°C}*5.70\°C\\\\Q_{nutrient}=4769.8 J[/tex]
Thus, in terms of the grams of the nutrient:
[tex]m_{nutrient}=0.0500mol*\frac{114g}{1mol}=5.70g[/tex]
The fuel value in nutritional Cal (kcal/g) turns out:
[tex]Fuel \ Value=\frac{-4.7698kJ}{5.70g}*\frac{1kcal}{4.184kJ}=200\frac{kcal}{g}[/tex]
Best regards!
