how much heat is needed to convert 826g of ice at -10.0 degrees celsius to steam at 126 degrees celsius
This is the same kind of problem as the previous one.
Change of state is mass x heat fusion/vaporization.
change of T within a state is
mass x specific heat x (Tfinal-Tinitial).
To determine the amount of heat needed to convert ice to steam, we must break down the process into different stages:
1. Heating the ice to its melting point.
2. Melting the ice.
3. Heating the resulting water from the melting point to its boiling point.
4. Vaporizing the water to steam.
Each stage requires a specific amount of heat, which can be calculated using the heat equation:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
Let's calculate the heat required for each stage:
1. Heating the ice to its melting point:
The specific heat capacity of ice is 2.09 J/g°C.
ΔT = (0°C - (-10.0°C)) = 10.0°C
Q1 = 826g * 2.09 J/g°C * 10.0°C
2. Melting the ice:
The heat of fusion (or latent heat of fusion) for ice is 333.55 J/g.
Q2 = 826g * 333.55 J/g
3. Heating the resulting water from the melting point to its boiling point:
The specific heat capacity of water is 4.18 J/g°C.
ΔT = (100.0°C - 0°C) = 100.0°C
Q3 = 826g * 4.18 J/g°C * 100.0°C
4. Vaporizing the water to steam:
The heat of vaporization (or latent heat of vaporization) for water is 2260 J/g.
Q4 = 826g * 2260 J/g
Now, to find the total heat required, we sum up the individual heat values:
Total heat = Q1 + Q2 + Q3 + Q4
Performing the calculations will give you the answer in joules.