4. A typical car engine produces 44,000 kJ of heat in an hour. The cooling system has a capacity of 8.40 L and is filled with a 50:50 mixture of anti-freeze (specific heat capacity of 8.37 J/g-oC, density of 1.038 g/mL). In maintaining a constant engine temperature, the cooling system temperature increased from 95oC to 110oC. (Assume this is within allowable limits.) How many times did the cooling system turn over its volume of anti-freeze to maintain the engine temperature during its hour of operation? Show your work. (Assume that only the cooling system removes heat from the engine and no other mechanism of heat removal is involved.)
My work so far:
First calculate mass of the coolant
8.7912 = 8.4 * 1.038.
Energy absorbed by coolant is
1,103.73516 = 8.7912 * 8.37 * 15
I can't figure out what to do next.
Well, don't fret! I'm here to help with a sprinkle of humor. Let's continue solving the problem together!
To find out how many times the cooling system turns over its volume of anti-freeze, we need to determine the volume of anti-freeze that the cooling system can heat by 15 degrees Celsius, which is how much the temperature increased. Exciting, right?
First, we need to calculate the specific heat capacity of the anti-freeze mixture in kJ/L-°C. To do that, we can convert the given value of 8.37 J/g-°C to kJ/L-°C by dividing it by 1000.
Now that's out of the way, let's move on to the equation:
Energy absorbed by the coolant = mass of coolant * specific heat capacity * temperature change
Since we already know the energy absorbed by the coolant is 1,103.73516 kJ, we can rearrange the equation to solve for mass:
Mass of coolant = energy absorbed by the coolant / (specific heat capacity * temperature change)
Plugging in the numbers, we get:
Mass of coolant = 1,103.73516 / (8.37 / 1000 * 15)
Now you can calculate the mass of the coolant!
Once we know the mass, we can figure out the volume of the coolant by dividing it by the density of 1.038 g/mL. Then, divide the volume of the coolant by the total volume of the cooling system (8.40 L) to find out how many times the system turns over its volume of anti-freeze.
Keep going with the calculations, and remember, math can be fun when you add a touch of clownery to it!
To find out how many times the cooling system turned over its volume of anti-freeze, you can utilize the concept of specific heat capacity and the change in temperature.
1. Calculate the heat absorbed by the coolant:
Heat = mass * specific heat capacity * temperature change
Heat = 8.7912 kg * 8.37 J/g-°C * (110°C - 95°C)
2. Convert the specific heat capacity from g to kg:
8.37 J/g-°C = 8.37 * 10^-3 J/kg-°C
3. Calculate the heat absorbed by the coolant:
Heat = 8.7912 kg * 8.37 * 10^-3 J/kg-°C * (110°C - 95°C)
4. Calculate the number of times the cooling system turned over its volume of anti-freeze:
Number of turnovers = Heat absorbed by the coolant / Energy produced by the engine
Number of turnovers = (8.7912 kg * 8.37 * 10^-3 J/kg-°C * (110°C - 95°C)) / 44000 kJ
5. Convert kJ to J:
Number of turnovers = (8.7912 kg * 8.37 * 10^-3 J/kg-°C * (110°C - 95°C)) / (44000 * 10^3 J)
Now you can perform the calculation to find the number of times the cooling system turned over its volume of anti-freeze.
To calculate the number of times the cooling system turns over its volume of anti-freeze, we need to determine the mass of the coolant that flows through the system in an hour.
First, let's calculate the heat energy absorbed by the coolant:
Energy absorbed by coolant = mass × specific heat capacity × temperature change
Given:
Specific heat capacity of anti-freeze (C): 8.37 J/g-°C
Initial temperature (T1): 95 °C
Final temperature (T2): 110 °C
To calculate the mass of the coolant, we can rearrange the formula as follows:
mass = energy absorbed by coolant / (specific heat capacity × temperature change)
mass = 1,103.73516 / (8.37 × (110 - 95))
mass ≈ 1,103.73516 / (8.37 × 15)
mass ≈ 8.7912 g
Now, let's calculate the volume of the coolant using the given density:
volume = mass / density
Note: The density provided for the anti-freeze is in g/mL, so we need to convert it to g/L:
density = 1.038 g/mL = 1.038 g/cm³
density = 1.038 g/cm³ × (1 mL/1 cm³) × (1000 cm³/1 L)
density ≈ 1038 g/L
volume = 8.7912 g / 1038 g/L
volume ≈ 0.00846 L
Now we can determine the number of times the coolant turns over its volume in an hour:
Number of turnovers = 44,000 kJ / energy absorbed per turnover
The energy absorbed per turnover is the energy absorbed by the coolant:
energy absorbed per turnover = 1,103.73516 J
Note: We convert 44,000 kJ to J by multiplying by 1000 (1 kJ = 1000 J).
Number of turnovers = 44,000,000 J / 1,103.73516 J
Number of turnovers ≈ 39,852
Therefore, the cooling system turned over its volume of anti-freeze approximately 39,852 times during its hour of operation.
I wouldn't tackled it this way, and you are free to not use what I am telling you because I can not be 100% positive about the answer. However, I would go this route:
q=mct
where
q=44,000 kJ=4.4 x 10^4 kJ=4.4 x 10^7 J
c=8.37 J/g*C
t=110 C-95 C=15 C
and
m=???
Solve for m:
m=q/c*t
m=(4.4 x 10^7 J)/[(8.37 J/g*C)(15 C)
You know m (mass in g), use the density to solve for the volume:
D=mass/volume(mL)
Mass/D=Volume (mL)
Convert mL to L:
1mL= 10^-3 L
Take the total volume, which I believe will be bigger, than the volume of the container, and solve for the number of cycles:
Volume that you solved for using the density and converted to L:
Volume in L/8.40L
***Answer with 3-significant figures.