If all of the chemical potential energy stored in a 100 Calorie snack pack was converted to thermal energy and subsequently absorbed by water, what would be the final temperature of the water? Assume you have a 1.5 mole sample of water initially at 25 degrees C. The specific heat of water is 4.184 J/g degrees C.
A 100 Calorie food snack pack actually is 100,000 "real" calories (the so-called big calorie is a kilocalorie). So,
q = mass x specific heat x (Tfinal-Tinitial).
You need to change 100,000 calories to joules (4.184 Joules = 1 calorie). You need to change 1.5 mole water to grams. Tfinal is the unknown and Tinitial is 25.
To find the final temperature of the water, we will use the principle of conservation of energy. The chemical potential energy stored in the snack pack is converted to thermal energy, which is then absorbed by the water. We can calculate the amount of thermal energy absorbed by the water using the equation:
Q = m * c * ΔT
Where:
Q is the thermal energy absorbed by the water (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (4.184 J/g degrees C)
ΔT is the change in temperature of the water (in degrees Celsius)
First, let's calculate the mass of the water using the given information that we have a 1.5 mole sample of water. The molar mass of water (H2O) is approximately 18 g/mol:
mass = 1.5 moles * 18 g/mol = 27 grams
Now, we can substitute the values into the formula and solve for ΔT:
Q = 100 Calories * 4.184 J/Calorie = 418.4 Joules
418.4 J = 27 g * 4.184 J/g °C * ΔT
ΔT = 418.4 J / (27 g * 4.184 J/g °C)
ΔT ≈ 4.06 °C
Therefore, the final temperature of the water would be 25 °C + 4.06 °C ≈ 29.06 °C.