If 25 grams of solid ice at-20 degrees C is heated to gaseous water at 125 degrees C, how much energy is required?
q1 = heat to warm ice from -20 to zero but not melt it.
q1 = mass ice x specific heat ice x delta T (delta T = 20).
q2 = heat to melt ice at zero.
q2 = mass ice x heat fusion.
q3 = heat to raise temperature from zero for liquid water to 100 C.
q3 = mass water x specific heat water x delta T (100 is delta T).
q4 = heat to change water from liquid to steam at 100.
q4 = mass water x heat vaporization.
q5 = heat to raise steam from 100 to 125 C.
q5 = mass steam x specific heat steam x delta T.
Total energy is q1 + q2 + q3 + q4 + q5.
You will need to look up specific heat ice, specific heat water, specific heat steam, heat of fusion and heat of vaporization.
Post your work if you get stuck.
To calculate the amount of energy required to heat a substance, we can use the following equation:
Q = mcΔT
Where:
Q is the energy required (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules/gram °C), and
ΔT is the change in temperature (in °C).
Let's break down the problem step by step:
Step 1: Determine the energy required to heat the ice from -20°C to its melting point (0°C).
Q1 = mcΔT
= 25g * 2.09 J/g °C * (0°C - (-20°C))
= 25g * 2.09 J/g °C * 20°C
= 1045 J
Step 2: Determine the energy required to melt the ice at 0°C.
The heat of fusion for water is 334 J/g.
Q2 = m * ΔHf
= 25g * 334 J/g
= 8350 J
Step 3: Determine the energy required to heat the liquid water from 0°C to 100°C.
Q3 = mcΔT
= 25g * 4.18 J/g °C * (100°C - 0°C)
= 25g * 4.18 J/g °C * 100°C
= 10450 J
Step 4: Determine the energy required to convert water from a liquid to a gas.
The heat of vaporization for water at 100°C is 2260 J/g.
Q4 = m * ΔHv
= 25g * 2260 J/g
= 56500 J
Step 5: Determine the energy required to heat the water vapor from 100°C to 125°C.
Q5 = mcΔT
= 25g * 2.09 J/g °C * (125°C - 100°C)
= 25g * 2.09 J/g °C * 25°C
= 1302.5 J
Step 6: Find the total energy required by adding up the energies from each step.
Total Energy = Q1 + Q2 + Q3 + Q4 + Q5
= 1045 J + 8350 J + 10450 J + 56500 J + 1302.5 J
= 79,647.5 J
Therefore, approximately 79,647.5 Joules of energy are required to heat 25 grams of solid ice at -20°C to gaseous water at 125°C.
To calculate the amount of energy required to heat the ice to gaseous water, we need to consider three different stages:
1. The energy required to raise the temperature of the ice from -20°C to 0°C.
2. The energy required to melt the ice at 0°C.
3. The energy required to heat the resulting liquid water from 0°C to 125°C.
Let's break down each stage and calculate the total energy required.
Stage 1: Raising the temperature of ice from -20°C to 0°C
The specific heat capacity of ice is 2.09 J/g°C. To calculate the energy required, we multiply the mass, temperature change, and specific heat capacity:
Energy = mass x temperature change x specific heat capacity
Energy = 25 g x (0 - (-20))°C x 2.09 J/g°C
Stage 2: Melting the ice at 0°C
The heat of fusion (or latent heat) for ice is 334 J/g. We need to multiply the mass of the ice by the heat of fusion:
Energy = 25 g x 334 J/g
Stage 3: Heating the liquid water from 0°C to 125°C
The specific heat capacity of liquid water is 4.18 J/g°C. Again, we multiply the mass, temperature change, and specific heat capacity:
Energy = mass x temperature change x specific heat capacity
Energy = 25 g x (125 - 0)°C x 4.18 J/g°C
Finally, to get the total energy required, we sum up the energy for each stage:
Total Energy Required = Energy (Stage 1) + Energy (Stage 2) + Energy (Stage 3)
Now you can substitute the values into the equations for each stage and calculate the total energy required.