In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 lb of oil provides 2.2×107J.

Part A
How many kilograms of oil are needed to heat 110 kg of water from 24 ∘C to 100∘C? (Assume that 1lb=454g, the specific heat of liquid water is 4.184 J/g∘C.)

To find out how many kilograms of oil are needed to heat 110 kg of water from 24°C to 100°C, we need to calculate the amount of thermal energy required to raise the temperature of the water.

The formula to calculate the thermal energy is:

Q = m * c * ΔT

where:
Q = thermal energy (in joules)
m = mass of the water (in kilograms)
c = specific heat capacity of water (in J/g°C or J/kg°C)
ΔT = change in temperature (in °C)

Let's break down the problem into steps.

Step 1: Convert the given mass of water from kg to grams.
110 kg = 110,000 grams

Step 2: Calculate the change in temperature.
ΔT = final temperature - initial temperature = 100°C - 24°C = 76°C

Step 3: Calculate the thermal energy.
Q = m * c * ΔT

The specific heat capacity of liquid water is 4.184 J/g°C. Since we want to use kilograms as the unit for mass, we'll convert the specific heat capacity from J/g°C to J/kg°C:
1 g = 0.001 kg

Thus, the specific heat capacity becomes 4.184 J/g°C * (1 g / 0.001 kg) = 4184 J/kg°C.

Now we can substitute the values into the formula:
Q = 110,000 g * 4184 J/kg°C * 76°C

Step 4: Calculate the thermal energy in joules.
Q = 110,000 g * 4184 J/kg°C * 76°C = 3.3 x 10^9 J

Finally, we need to calculate the mass of oil needed to produce this amount of thermal energy.

Given that the combustion of 1.0 lb (454 g) of oil provides 2.2 x 10^7 J of energy, we can calculate the mass of oil required using the following equation:

M_oil = Q / Energy_per_mass

where:
M_oil = mass of oil (in kilograms)
Q = thermal energy (in joules)
Energy_per_mass = energy provided by 1 lb (454 g) of oil (in joules)

Substituting the values:
M_oil = 3.3 x 10^9 J / 2.2 x 10^7 J/kg

M_oil = 150 kg

Therefore, 150 kilograms of oil are needed to heat 110 kilograms of water from 24°C to 100°C.

To find out how many kilograms of oil are needed to heat 110 kg of water from 24 ∘C to 100∘C, we need to calculate the heat energy required to heat the water and then convert it into the equivalent amount of oil.

Step 1: Calculate the heat energy required to heat the water.

The heat energy required to heat a substance is given by the formula:

Q = m * c * ΔT

Where:
Q = heat energy (in Joules)
m = mass of the substance (in kilograms)
c = specific heat capacity of the substance (in J/g∘C)
ΔT = change in temperature (in ∘C)

Given:
Mass of water (m) = 110 kg
Change in temperature (ΔT) = 100∘C - 24 ∘C = 76 ∘C
Specific heat capacity of water (c) = 4.184 J/g∘C

Step 2: Convert the mass of water from kilograms to grams.

1 kg = 1000 g

So, 110 kg = 110 * 1000 g = 110000 g

Step 3: Calculate the heat energy required.

Q = m * c * ΔT
= 110000 g * 4.184 J/g∘C * 76 ∘C

Step 4: Convert the heat energy from Joules to pounds.

Given:
1 lb = 454 g
1 J = 1/2.2×107 lb

So, the heat energy in pounds is:

Q_lb = Q * (1 lb / 454 g) * (1 J / (2.2×107 lb))

Step 5: Convert the heat energy in pounds to kilograms of oil.

Since we know that the combustion of 1.0 lb of oil provides 2.2×107 J, we can set up a proportion to find the equivalent amount of oil:

Q_lb / 1 lb = q_oil / 2.2×107 J

Simplifying the proportion:

q_oil = Q_lb * 2.2×107 J / 1 lb

Now, we can substitute the value of Q_lb to find the amount of oil in kilograms:

q_oil = (Q * (1 lb / 454 g) * (1 J / (2.2×107 lb))) * 2.2×107 J / 1 lb

Finally, simplify the equation and calculate:

q_oil = Q * 2.2×107 J / (454 g * 2.2×107 lb)
= Q / 454 grams
= 110000 g * 4.184 J/g∘C * 76 ∘C / 454 grams

After performing the calculation, the answer is the amount of oil in kilograms, q_oil.

q needed to heat the water is

q = mass H2O x specific heat H2O x (Tfinal-Tinitial). Note that if you use specific heat listed you must change 110 kg H2O to grams.

1 lb x (? J needed/2.2E7 J/lb) = lbs oil needed. Convert to kg oil.