At 1 atm, how much energy is required to heat 37.0g of H2O at -24.0 C to H2O at 155.0 C?
q1 = heat needed to heat ice at -24 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tf is zero and Ti is -24
q2 = heat needed to change solid ice at zero to liquid water at zero.
q2 = mass ice x heat fusion.
q3 = heat needed to heat water at zero to water at 100 C.
q3 = mass water x specific heat H2O x (Tfinal-Tinitial) where Tf is 100 and Ti = 0.
q4 = heat needed to change liquid water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization
q5 = heat needed to heat steam and raise temperature from 100 C to 155 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial) where Tf is 155 and Ti is 100.
Total Q = sum of each q; i.e.,
Qtotal = q1 + q2 + q3 + q4 + q5
To calculate the energy required to heat a substance, you can use the equation:
q = m * c * ΔT
Where:
q is the energy (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in joules/gram °C)
ΔT is the change in temperature (in °C)
In this case, the substance is water (H2O), and we need to find the energy required to heat 37.0g of water from -24.0°C to 155.0°C at a constant pressure of 1 atm.
First, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Next, we calculate the change in temperature:
ΔT = final temperature - initial temperature
= 155.0°C - (-24.0°C)
= 179.0°C
Now, substitute the given values into the equation:
q = m * c * ΔT
= 37.0 g * 4.18 J/g°C * 179.0°C
≈ 27,346.06 J
Therefore, the energy required to heat 37.0g of water from -24.0°C to 155.0°C at 1 atm pressure is approximately 27,346.06 Joules.