a 100 g sample of ice at -15 degrees celsius is heated until it is converted to water vapour at 120 degrees celsius what is the energy change?
q1 = heat needed to raise T of solid ice from -15 to zero C.
q1 = mass ice x specific heat x (Tfinal-Tinitial)
q2 = heat to melt ice at zero C to liquid H2O at zero C.
q2 = mass ice x heat fusion
q3 = heat needed to raise T of liquid water from zero C to 100 C.
q3 = mass water x specific heat H2O x (Tfinal-Tinitial)
q4 = heat needed to convert liquid water @ 100 C to vapor at 100 C
q4 = mass H2O x heat vaporization
q5 = heat needed to raise T of steam @ 100 C to steam at 120 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = q1 + q2 + q3 + q4 + q5
Still don't understand this! Please help!
To find the energy change, we need to consider the different phases the substance goes through and the associated energy changes. Let's break down the steps:
Step 1: Heating the ice from -15 degrees Celsius to 0 degrees Celsius
The energy change in this step is given by the formula:
q1 = m * c * ΔT
where:
q1 = energy change
m = mass of the ice (100 g)
c = specific heat capacity of ice (2.09 J/g°C)
ΔT = change in temperature (0°C - (-15°C) = 15°C)
Plugging in the values:
q1 = 100 g * 2.09 J/g°C * 15°C
q1 = 3135 J
Step 2: Melting the ice at 0 degrees Celsius
The energy change during this step is called the heat of fusion. The heat of fusion for ice is 334 J/g.
q2 = m * ΔHfus
where:
q2 = energy change
m = mass of the ice (100 g)
ΔHfus = heat of fusion (334 J/g)
Plugging in the values:
q2 = 100 g * 334 J/g
q2 = 33400 J
Step 3: Heating the water from 0 degrees Celsius to 100 degrees Celsius
The energy change in this step is given by the formula:
q3 = m * c * ΔT
where:
q3 = energy change
m = mass of the water (100 g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (100°C - 0°C = 100°C)
Plugging in the values:
q3 = 100 g * 4.18 J/g°C * 100°C
q3 = 41800 J
Step 4: Vapourizing the water at 100 degrees Celsius
The energy change during this step is called the heat of vaporization. The heat of vaporization for water is 2260 J/g.
q4 = m * ΔHvap
where:
q4 = energy change
m = mass of the water (100 g)
ΔHvap = heat of vaporization (2260 J/g)
Plugging in the values:
q4 = 100 g * 2260 J/g
q4 = 226000 J
Step 5: Heating the water vapor from 100 degrees Celsius to 120 degrees Celsius
The energy change in this step is given by the formula:
q5 = m * c * ΔT
where:
q5 = energy change
m = mass of the water vapor (100 g)
c = specific heat capacity of water vapor (2.03 J/g°C)
ΔT = change in temperature (120°C - 100°C = 20°C)
Plugging in the values:
q5 = 100 g * 2.03 J/g°C * 20°C
q5 = 4060 J
Total Energy Change:
The total energy change is the sum of all the energy changes:
Total energy change = q1 + q2 + q3 + q4 + q5
Total energy change = 3135 J + 33400 J + 41800 J + 226000 J + 4060 J
Total energy change = 308395 J
Therefore, the energy change when the 100 g sample of ice at -15 degrees Celsius is converted to water vapor at 120 degrees Celsius is 308,395 Joules.