1.10 g of glucose was completely burnt in a copper calorimeter. The temperature of the water
increased from 25.85 °C to 36.50 °C.
You would have to find the Enthaply change of the water (just use Q=mcΔΤ) and use it as a negative value as it is the energy taken from the glucose reaction. Then by dividing with the number of moles that 1.1 g of glucose has (hint n=m/M) you would be able to find the joules per mol. If I am not mistaken..
Urbmom
yep what danAI said
To find the heat absorbed by the water, you can use the formula:
Q = mcΔT
Where:
Q = Heat absorbed by the water (in Joules)
m = Mass of water (in grams)
c = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature (in °C)
First, you need to find the mass of the water. This can be done by assuming that the density of water is 1 g/mL, so 1 mL of water has a mass of 1 gram. Assuming that the density of the water remains constant during the experiment, you can assume that the mass of water is equal to the volume of water used. However, you need to consider that some volume of water may have evaporated during the experiment, so you need to account for that.
Let's assume that 1 mL of water corresponds to 1 gram of water initially. If the water temperature increased by ΔT, you can calculate the mass of water as follows:
Mass of water = Volume of water × Density of water
= ΔT × 1 g/mL
Now, let's calculate the mass of water:
ΔT = 36.50 °C - 25.85 °C
= 10.65 °C
Mass of water = 10.65 g
Next, you can substitute the known values into the formula to find the heat absorbed by the water:
Q = mcΔT
= (10.65 g) × (4.18 J/g°C) × (10.65 °C)
Finally, calculate the heat absorbed by the water:
Q ≈ 450.067 Joules
Therefore, the heat absorbed by the water is approximately 450.067 Joules.