How much energy would it take to warm 30.0g of water at 35 C to the boiling point and boil all the liquid away?
I don't believe your question has been answered. Here is what I would do.
q1 = mcdT.
q1= ?
m = 30 grams H2O
c = specific heat. For water that is 4.184 J/g*C
dT = change in T = 100-35 = ?
Using this information solve for q1.
To boil the water.
q2 = mass x heat vaporization.
q2 = 30 g x 2260 J/g = ?
Total energy = q1 + q2 = ?
Convert to kJ if needed
Well, let's see. To warm up the water from 35°C to its boiling point, we would need to provide it with some energy. This can be calculated using the specific heat capacity of water, which is about 4.18 J/g°C.
So, the energy required to warm up the water is:
30.0 g × 4.18 J/g°C × (100°C - 35°C) = 30,450 J
Now, to boil the water, we need to consider the heat of vaporization of water, which is around 2260 J/g.
The energy required to boil all the liquid away is:
30.0 g × 2260 J/g = 67,800 J
Adding these two energy values together, we get a total energy requirement of:
30,450 J + 67,800 J = 98,250 J
So, it would take approximately 98,250 joules of energy to warm and boil away the 30.0g of water. That's quite a bit of energy, but don't worry, we won't let it boil your brain!
To calculate the amount of energy required to heat and evaporate water, we need to consider two separate processes: heating the water to its boiling point and then vaporizing it.
1. Heating the water to its boiling point:
The specific heat capacity of water is approximately 4.18 J/g°C. This means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Given:
Mass of water (m) = 30.0 grams
Change in temperature (ΔT) = (100°C - 35°C) = 65°C
To calculate the energy required to heat the water, we can use the formula:
Energy = mass × specific heat capacity × temperature change
Energy = 30.0 g × 4.18 J/g°C × 65°C
Energy = 8,157 Joules
Therefore, it would take 8,157 Joules of energy to heat the 30.0 grams of water to the boiling point.
2. Vaporizing the water:
The heat of vaporization for water is approximately 2.26 × 10^3 J/g. This means it takes 2.26 × 10^3 Joules of energy to convert 1 gram of water from a liquid to a gas at its boiling point.
Given:
Mass of water (m) = 30.0 grams
To calculate the energy required to boil away the water, we can use the formula:
Energy = mass × heat of vaporization
Energy = 30.0 g × 2.26 × 10^3 J/g
Energy = 67,800 Joules
Therefore, it would take 67,800 Joules of energy to boil away the 30.0 grams of water.
Adding the energy required for heating and boiling away the water:
Total energy = Energy for heating + Energy for boiling
Total energy = 8,157 Joules + 67,800 Joules
Total energy = 75,957 Joules
Therefore, it would take a total of 75,957 Joules of energy to warm 30.0 grams of water at 35°C to its boiling point and then evaporate all the liquid away.
To determine the energy required to warm and boil water, we need to consider two separate steps:
Step 1: Calculate the energy required to raise the temperature of water from 35°C to its boiling point.
Step 2: Calculate the energy required to convert the water at its boiling point to steam.
Let's break down each step and calculate the energy required:
Step 1: Calculate the energy required to raise the temperature of water from 35°C to its boiling point.
To calculate the energy required to raise the temperature of a substance, we use the formula:
Q = m * c * ΔT
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 J/g°C),
ΔT is the change in temperature (in °C).
For water:
m = 30.0g (given)
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = (100°C - 35°C) = 65°C
Now, let's substitute the values into the formula:
Q = 30.0g * 4.18 J/g°C * 65°C = 8,157 J
Therefore, it would require 8,157 Joules of energy to raise the temperature of 30.0 grams of water from 35°C to its boiling point.
Step 2: Calculate the energy required to convert the water at its boiling point to steam.
To calculate the energy required to convert water from its boiling point to steam, we use the formula:
Q = m * ΔH_vap
Where:
Q is the energy required (in Joules),
m is the mass of water (in grams),
ΔH_vap is the heat of vaporization for water (in J/g).
For water:
m = 30.0g (given)
ΔH_vap = 40.7 J/g (heat of vaporization for water)
Now, let's substitute the values into the formula:
Q = 30.0g * 40.7 J/g = 1,221 J
Therefore, it would require 1,221 Joules of energy to convert 30.0 grams of water at its boiling point to steam.
To find the total energy required, we need to add the energy from both steps:
Total Energy = Step 1 energy + Step 2 energy
Total Energy = 8,157 J + 1,221 J = 9,378 J
Therefore, it would require approximately 9,378 Joules of energy to warm 30.0 grams of water at 35°C to its boiling point and boil all the liquid away.