calculate the heat released when 166g of water at 23.0 degrees celsius are converted to ice at -10.0 degrees celsius
q1 = heat released to move water @ 23C to zero C.
q1 = mass water x specific heat water x (Tfinal-Tinitial).
q2 = heat released to freeze water @ zero C to ice @ zero C.
q2 = mass water x heat fusion.
q3 = heat released to move ice from zero C to -10C.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial).
Total Q = q1 + q2 + q3
322.5
To calculate the heat released during the conversion of water to ice, you need to use the equation:
q = m * ΔH
Where:
q is the heat released or absorbed
m is the mass of the substance
ΔH is the enthalpy change
First, calculate the heat released during the cooling of water from 23.0°C to 0°C:
q1 = m * C * ΔT1
Where:
m is the mass of water
C is the specific heat capacity of water (4.18 J/g°C)
ΔT1 is the change in temperature (23.0°C - 0°C)
q1 = 166 g * 4.18 J/g°C * (23.0°C - 0°C)
q1 = 166 g * 4.18 J/g°C * 23.0°C
q1 = 18142.68 J
Next, calculate the energy released during the phase change from water to ice at 0°C:
q2 = m * ΔHf
Where:
m is the mass of water
ΔHf is the enthalpy of fusion for water (334 J/g)
q2 = 166 g * 334 J/g
q2 = 55444 J
Then, calculate the heat released during the cooling of ice from 0°C to -10°C:
q3 = m * C * ΔT2
Where:
m is the mass of ice
C is the specific heat capacity of ice (2.09 J/g°C)
ΔT2 is the change in temperature (0°C - (-10.0°C))
q3 = 166 g * 2.09 J/g°C * (0°C - (-10.0°C))
q3 = 166 g * 2.09 J/g°C * 10.0°C
q3 = 3465.4 J
Finally, add up all the individual heat values to find the total heat released:
q_total = q1 + q2 + q3
q_total = 18142.68 J + 55444 J + 3465.4 J
q_total = 77051.08 J
Therefore, the heat released when 166g of water at 23.0 degrees Celsius is converted to ice at -10.0 degrees Celsius is approximately 77051.08 J.