How much heat (in g ) is evolved in converting 1.00 of steam at 145.0 to ice at -45.0? The heat capacity of steam is 2.01 and of ice is 2.09 .
q1 = heat removed to convert steam at 145 C to steam at 100 C.
q1 = mass x specific heat steam x (Tfinal-Tinitial)-----where Tf is 100 and Ti = 145.
q2 = heat removed to convert steam at 100 to water at 100 C.
q2 = mass steam x heat vaporization
q3 = heat removed to convert water at 100 to water at 0 C.
q3 = mass water x specific heat water x delta T.
q4 = heat removed to convert water at 0 C to ice at 0 C.
q4 = mass x heat fusion water.
q5 = heat removed to convert ice at 0C to -45 C.
q5 = mass ice x specific heat ice x delta T.
Total heat removed (evolved) = q1 + q2 + q3 + q4 + q5
57.6
To calculate the heat evolved in converting steam to ice, we can divide the process into two steps:
1. Heating steam from 145.0 to 0 degrees Celsius
2. Cooling steam from 0 degrees Celsius to -45 degrees Celsius and then converting it to ice.
Step 1: Heating steam from 145.0 to 0 degrees Celsius
To calculate the heat evolved in this step, we can use the formula:
q1 = mass * heat capacity * temperature change
Given:
Mass of steam = 1.00 g
Heat capacity of steam = 2.01 g/°C
Temperature change = 0 - 145.0 = -145.0 °C
q1 = 1.00 g * 2.01 g/°C * (-145.0 °C)
q1 = -292.45 g°C
Step 2: Cooling steam from 0 degrees Celsius to -45 degrees Celsius and converting it to ice
To calculate the heat evolved in this step, we can use the formula:
q2 = mass * heat capacity * temperature change
Given:
Mass of steam = 1.00 g
Heat capacity of ice = 2.09 g/°C
Temperature change = -45.0 °C - 0 °C = -45.0 °C
q2 = 1.00 g * 2.09 g/°C * (-45.0 °C)
q2 = -94.05 g°C
The total heat evolved in converting steam to ice is the sum of q1 and q2:
Total heat evolved = q1 + q2
Total heat evolved = -292.45 g°C + -94.05 g°C
Total heat evolved = -386.5 g°C
Therefore, the heat evolved in converting 1.00 g of steam at 145.0°C to ice at -45.0°C is -386.5 g°C.
To calculate the amount of heat evolved in converting steam to ice, we need to consider the heat energy absorbed or released during each phase change.
The process involves three steps:
1. Cooling the steam from 145.0°C to 100.0°C at constant pressure.
2. Condensing the steam at 100.0°C to liquid water at 100.0°C.
3. Cooling the liquid water from 100.0°C to -45.0°C and converting it to ice.
Step 1: Cooling the steam from 145.0°C to 100.0°C
To calculate the heat absorbed during this step, we can use the formula:
Q = m * C * ΔT
where:
Q = heat energy
m = mass of the substance (in grams)
C = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Given:
mass of steam (m) = 1.00 g
specific heat capacity of steam (C) = 2.01 J/g°C
change in temperature (ΔT) = 100.0°C - 145.0°C = -45.0°C
Q1 = 1.00 g * 2.01 J/g°C * (-45.0°C)
Step 2: Condensing the steam at 100.0°C to liquid water at 100.0°C
During this phase change, the heat energy is given by the formula:
Q = m * ΔH_vap
where:
Q = heat energy
m = mass of the substance (in grams)
ΔH_vap = heat of vaporization (in J/g)
Given that the heat of vaporization of steam is 40.7 J/g, we can calculate Q2 as:
Q2 = 1.00 g * 40.7 J/g
Step 3: Cooling the liquid water from 100.0°C to -45.0°C and converting it to ice
Using the same formula as in step 1, we can calculate the heat absorbed during this step.
Given:
mass of liquid water (m) = 1.00 g
specific heat capacity of ice (C) = 2.09 J/g°C
change in temperature (ΔT) = -45.0°C - 100.0°C = -145.0°C
Q3 = 1.00 g * 2.09 J/g°C * (-145.0°C)
The total heat evolved during the conversion is given by:
Total heat evolved = Q1 + Q2 + Q3
Therefore, calculate the values for Q1, Q2, Q3, and sum them up to determine the amount of heat evolved in converting 1.00 g of steam at 145.0°C to ice at -45.0°C.