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Sagot :
We must first determine the heat needed to heat 175mL of water. In the working temperature range we can assume that the density of water is 1g/mL, so the mass of water that we must heat will be 175g of water.
Now, the heat we need to heat that mass of water is calculated with the following equation:
[tex]Q=mCp\Delta T[/tex]Where Q is the heat required to heat the water
Cp is the specific heat of water equal to 4186 J/g°C
delta T is the change of temperature = 25°C-20°C=5°C
We replace the know values:
[tex]Q=175g\times4186\frac{J}{g\degree C}\times\frac{1kJ}{1000J}\times5\degree C=3662.75kJ[/tex]Now, to heat the water they tell us that the glucose combustion will take place. We must assume that all the heat generated in the reaction will go to heating the water and that there is no heat loss to the environment.
The combustion reaction of glucose generates 2538.7 kJ/mol. So the moles of glucose needed will be:
[tex]\begin{gathered} molGlucose=givenHeatrequired\times\frac{1molGlucose}{2538.7kJ} \\ molGlucose=3662.75kJ\times\frac{1molGlucose}{2,538.7kJ}=1.44molGlucose \end{gathered}[/tex]We calculate the grams of glucose by multiplying the moles by the molar mass of glucose equal to 180.2g/mol
grams of glucose= 1.44 mol x 180.2 g/mol=260g
Answer: To heat 175 mL of water from 20.0 °C to 25.0°C are required 260g of glucose
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