How much heat energy is needed to change the temperature of 25.00 mL of water from 5.0°C to 75.0°C? The molar heat capacity of water for this process Cp = 75.38 J/(mol °C), d = 1.00 g/mL.
q = mols H2O x molar heat capacity x (Tfinal-Tinitial)
To calculate the amount of heat energy needed to change the temperature of water, we need to use the equation:
q = m * Cp * ΔT
Where:
q is the heat energy (in joules),
m is the mass of the water (in grams),
Cp is the molar heat capacity of water (in joules per mole per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
First, we need to convert the volume of water from milliliters to grams, using the density (d):
m = V * d
Given:
V = 25.00 mL
d = 1.00 g/mL
m = 25.00 mL * 1.00 g/mL
m = 25.00 g
Now we can calculate the heat energy (q):
ΔT = final temperature - initial temperature
ΔT = 75.0°C - 5.0°C
ΔT = 70.0°C
q = 25.00 g * Cp * 70.0°C
Given:
Cp = 75.38 J/(mol °C)
We need to convert grams to moles using the molar mass of water:
Molar mass of water (H₂O) = 2(1.01 g/mol) + 16.00 g/mol
Molar mass of water (H₂O) ≈ 18.02 g/mol
moles = mass / molar mass
moles = 25.00 g / 18.02 g/mol
Now we can calculate q:
q = (25.00 g / 18.02 g/mol) * 75.38 J/(mol °C) * 70.0°C
By substituting the given values into the equation, we can calculate the heat energy (q).