How much heat (in ) is evolved in converting 1.00 of steam at 150.0 to ice at -55.0? The heat capacity of steam is 2.01 and of ice is 2.09 .
To find the amount of heat evolved in converting steam to ice, we need to calculate the heat exchanged during the process.
The heat transferred during a phase change can be calculated using the equation:
Q = m * ∆H
where:
Q is the heat transferred
m is the mass of the substance
∆H is the heat of transformation (also known as the enthalpy change)
In this case, we are converting 1.00 g of steam at 150.0 °C to ice at -55.0 °C.
Step 1: Calculate the heat transferred during the phase change from steam to ice using the equation above.
For steam:
Q1 = m * ∆Hvap
where ∆Hvap is the heat of vaporization for water. The heat of vaporization for water is 40.7 kJ/mol, which is equivalent to 40.7 J/g.
For ice:
Q2 = m * ∆Hfus
where ∆Hfus is the heat of fusion for ice. The heat of fusion for ice is 6.01 kJ/mol, which is equivalent to 6.01 J/g.
Step 2: Calculate the mass of the substance using the given values.
The mass is given as 1.00 g.
Step 3: Calculate the heat exchanged during the process.
Q1 = m * ∆Hvap
Q1 = 1.00 g * 40.7 J/g
Q1 = 40.7 J
Q2 = m * ∆Hfus
Q2 = 1.00 g * 6.01 J/g
Q2 = 6.01 J
Step 4: Add the heat transferred during the phase change from steam to ice.
Q = Q1 + Q2
Q = 40.7 J + 6.01 J
Q ≈ 46.71 J
Therefore, approximately 46.71 Joules of heat are evolved in converting 1.00 g of steam at 150.0 °C to ice at -55.0 °C.
add the heats.
cooling steam to 100: mass*specheatsteam*50
converting to water:
mass*heatvaporizationwater
cooling water:
mass*specificheatwater*(100)
freezing water
mass*Heatfusion
cooling ice
mass*speciheatice*(55)
add.