calculate the enthalpy change in delta H foor heating 250 grams of liquid water from 0 degrees celcius to 100 degrees celcius.
q = mass x specific heat x delta T.
Note the correct spelling of celsius
To calculate the enthalpy change (ΔH) for heating 250 grams of liquid water from 0 degrees Celsius to 100 degrees Celsius, you need to use the equation:
ΔH = m * C * ΔT
Where:
ΔH = Enthalpy change (in joules)
m = Mass of the substance (in grams)
C = Specific heat capacity of the substance (in joules per gram per degree Celsius)
ΔT = Change in temperature (in degrees Celsius)
For water, the specific heat capacity is approximately 4.18 J/g°C.
First, calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 100°C - 0°C
ΔT = 100°C
Now, substitute the given values into the equation:
ΔH = 250 g * 4.18 J/g°C * 100°C
The unit "g" cancels out, giving you the resulting unit in joules (J):
ΔH = 250 * 4.18 * 100 J
Solve the equation:
ΔH = 104,500 J (or 104.5 kJ)
Therefore, the enthalpy change for heating 250 grams of liquid water from 0 degrees Celsius to 100 degrees Celsius is 104,500 Joules (or 104.5 kilojoules).