In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 pound of oil provides 2.4*10^7 J.
How many kg of oil are needed to heat 100 kg of water from 22 to 100 degrees C?
How many kg of oil are needed to provide steam from 100 kg of water at 100 degrees C?
a. 100 kg H2O from 22 to 100 will require how much energy?
That is mass H2O (in grams) x specific heat H2O (J/g*C) x (100-22) = ?J
oil needed(lbs) = 1 lb x (?J/2.4E7)
Convert lbs oil to kg.
b. Same concept but heat needed is
100,000 g H2O x delta Hvap = heat needed.
To find out how many kilograms of oil are needed to heat 100 kg of water from 22 to 100 degrees C, we need to calculate the amount of energy required.
Step 1: Calculate the heat energy required to raise the temperature of water from 22 to 100 degrees C.
The specific heat capacity of water is approximately 4186 J/kg°C.
The temperature change is 100 - 22 = 78 degrees C.
Heat energy = mass * specific heat capacity * temperature change
Heat energy = 100 kg * 4186 J/kg°C * 78 °C
Step 2: Convert the heat energy from joules to pounds.
1 pound of oil provides 2.4*10^7 J.
So, the energy required in pounds = heat energy / (2.4*10^7 J/pound)
Step 3: Convert the energy required in pounds to kilograms.
1 pound = 0.4536 kg.
So, the energy required in kilograms = energy required in pounds * (0.4536 kg/pound)
Now let’s calculate:
Step 1:
Heat energy = 100 kg * 4186 J/kg°C * 78 °C = 3,265,200 J
Step 2:
Energy required in pounds = 3,265,200 J / (2.4*10^7 J/pound) ≈ 0.1361 pounds
Step 3:
Energy required in kilograms = 0.1361 pounds * (0.4536 kg/pound) = 0.0618 kg
Therefore, approximately 0.0618 kg of oil is needed to heat 100 kg of water from 22 to 100 degrees C.
Now, let's move on to the second question:
To find out how many kilograms of oil are needed to provide steam from 100 kg of water at 100 degrees C, we need to calculate the heat energy required for vaporization.
Step 1: Calculate the heat energy required to vaporize 100 kg of water.
The latent heat of vaporization of water is approximately 2.26*10^6 J/kg.
Heat energy = mass * latent heat of vaporization
Heat energy = 100 kg * 2.26*10^6 J/kg
Step 2: Convert the heat energy from joules to pounds and then to kilograms (if necessary).
1 pound of oil provides 2.4*10^7 J.
Now let’s calculate:
Step 1:
Heat energy = 100 kg * 2.26*10^6 J/kg = 2.26*10^8 J
Step 2:
Energy required in pounds = 2.26*10^8 J / (2.4*10^7 J/pound) ≈ 9.41 pounds
Energy required in kilograms = 9.41 pounds * (0.4536 kg/pound) ≈ 4.26 kg
Therefore, approximately 4.26 kg of oil is needed to provide steam from 100 kg of water at 100 degrees C.
To solve these problems, we need to use the concept of specific heat capacity and latent heat of vaporization.
First, let's address the first question: How many kilograms of oil are needed to heat 100 kg of water from 22 to 100 degrees Celsius?
To find this, we need to calculate the amount of heat energy required to raise the temperature of the water. We can use the formula:
Q = mcΔT
where:
Q is the amount of heat energy,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C or 4180 J/kg°C.
Let's calculate step by step:
1. Calculate the mass of water required to be heated. Given that the mass of water is 100 kg.
2. Calculate the change in temperature. ΔT = (final temperature - initial temperature) = (100°C - 22°C) = 78°C.
3. Now, calculate the amount of heat energy required using the formula Q = mcΔT:
Q = (mass of water) * (specific heat capacity of water) * (change in temperature)
Q = (100 kg) * (4180 J/kg°C) * (78°C)
Finally, we can calculate how many kilograms of oil are needed to produce this amount of heat energy.
4. Given that the combustion of 1.0 pound (approximately 0.454 kg) of oil provides 2.4 * 10^7 J, we can calculate the number of kilograms of oil required using the following formula:
(mass of oil) * (heat energy per kg of oil) = (amount of heat energy required)
Let's substitute the values:
(mass of oil) * (2.4 * 10^7 J/kg) = (Q)
Now, solve for the mass of oil:
mass of oil = (Q) / (2.4 * 10^7 J/kg)
Substitute the calculated value of Q and solve for the mass of oil.
Now let's move on to the second question: How many kilograms of oil are needed to provide steam from 100 kg of water at 100 degrees Celsius?
To find this, we need to calculate the latent heat of vaporization of water.
The latent heat of vaporization is the amount of heat energy required to change 1 kg of water from a liquid to a gas at its boiling point.
The latent heat of vaporization for water is approximately 2.26 * 10^6 J/kg.
Using a similar approach as before, we can calculate the mass of oil required to produce this amount of heat energy.
(mass of oil) * (heat energy per kg of oil) = (amount of heat energy required)
Substitute the values:
(mass of oil) * (2.4 * 10^7 J/kg) = (latent heat of vaporization)
Solve for the mass of oil using the given value of the latent heat of vaporization.