The temperature of 100.0 mL of water was raised from 26.7 degrees C to 62.3 degrees C. How much heat in calories was added to the water? (The specific heat of water is 1.000 cal/g degrees C and the density of water at 26.7 degrees C is 0.996626 g/mL)
q = mass water x specific heat water x delta T.
3560cal
To determine how much heat was added to the water, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat added or gained by the water (in calories),
m is the mass of the water (in grams),
c is the specific heat of water (in cal/g degrees C),
ΔT is the change in temperature (in degrees C).
First, let's find the mass of the water. We know that the volume of the water is 100.0 mL, and the density of water at 26.7 degrees C is 0.996626 g/mL. We can calculate the mass using the formula:
mass = volume * density
mass = 100.0 mL * 0.996626 g/mL
mass = 99.6626 g
Now, we have the mass of the water as 99.6626 g.
Next, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 62.3 degrees C - 26.7 degrees C
ΔT = 35.6 degrees C
Now, we have the mass of the water (m = 99.6626 g) and the change in temperature (ΔT = 35.6 degrees C).
Finally, we can substitute these values into the formula to find the amount of heat added (Q):
Q = m * c * ΔT
Q = 99.6626 g * 1.000 cal/g degrees C * 35.6 degrees C
Q = 3560.0316 cal
Therefore, the amount of heat added to the water is approximately 3560.03 calories.