how much heat is required to warm 225 grams of water from -50c to 200 c
q1 = heat needed to raise solid ice from -50 to zero C.
q1 = mass H2O x specific heat ice x (Tfinal-Tinitial)
q2 = heat needed to melt solid ice at zero C to liquid water at zero C.
q2 = mass H2O x heat fusion
q3 = heat needed to raise T of liquid water from zero C to 100 C.
q3 = mass H2O x specific heat water x (Tfinal-Tinitial)
q4 = heat needed to convert H2O at 100 C to steam at 100 C.
q4 = mass H2O x heat vaporization
q5 = heat needed to raise T of steam at 100 to steam at 200 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = sum of individual steps.
Post your work in detail if you get stuck.
To calculate the amount of heat required to warm a substance, you need to use the specific heat capacity formula, which is:
Q = m * c * ΔT
Where:
Q = Heat energy (in Joules)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in J/g°C)
ΔT = Change in temperature (in °C)
First, let's calculate the temperature change (ΔT):
ΔT = Final temperature - Initial temperature
= 200°C - (-50°C)
= 250°C
Now, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Plugging in the values into the formula:
Q = 225g * 4.18 J/g°C * 250°C
Q ≈ 23,512.5 Joules
Therefore, approximately 23,512.5 Joules of heat energy is required to warm 225 grams of water from -50°C to 200°C.