Check my work & answer please. I need to make sure I am able to do this type of problem correctly.
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:
Q = mct
44000000 = m*8.37*15
M = 350,457.984866587
D=m/v
1.038 = 350,457.984866587/v
Mass/d = volume
350,457.984866587/1.038 = 337,628.116442ml
337.628116442L
40,193.8233859 turns
That looks ok to me but you're not allowed that many significant figures. If it were me I would round to 40.2 times.
To check your work, let's go through the problem step-by-step:
1. Calculate the heat transferred from the engine to the cooling system:
Q = m * c * ΔT
Given: Q = 44,000 kJ, c = 8.37 J/g-°C, ΔT = 110°C - 95°C = 15°C
44,000 kJ = m * 8.37 J/g-°C * 15°C
44,000,000 J = m * 8.37 J/g-°C * 15°C
Dividing both sides by 8.37 J/g-°C * 15°C, we get:
m = 44,000,000 J / (8.37 J/g-°C * 15°C)
m ≈ 349,941.87 g
2. Calculate the volume of the antifreeze in the cooling system:
V = m / D
Given: D = 1.038 g/mL
V = 349,941.87 g / 1.038 g/mL
V ≈ 336,857.03 mL
Converting mL to L, we get:
V ≈ 336.85703 L
3. Determine the number of times the cooling system turned over its volume of antifreeze:
Number of turnovers = Volume of antifreeze flow / Volume of the cooling system
Given: Volume of the cooling system = 8.40 L
Number of turnovers = 336.85703 L / 8.40 L
Number of turnovers ≈ 40.08
Therefore, the cooling system turned over its volume of antifreeze approximately 40 times during its hour of operation.
Your answer of 40,193.8233859 turns seems incorrect. Please double-check your calculations and make sure you took the correct values and units into account.
To check your work, let's go through the problem and calculate the correct answer step by step.
Given:
- Heat produced by the car engine in an hour: 44,000 kJ
- Capacity of the cooling system: 8.40 L
- Cooling system filled with a 50:50 mixture of antifreeze
- Specific heat capacity of antifreeze: 8.37 J/g-°C
- Density of antifreeze: 1.038 g/mL
- Initial temperature of the cooling system: 95°C
- Final temperature of the cooling system: 110°C
Step 1: Calculate the change in temperature of the cooling system.
ΔT = Final temperature - Initial temperature
ΔT = 110°C - 95°C
ΔT = 15°C
Step 2: Calculate the total amount of heat absorbed by the cooling system.
Q = m * c * ΔT
Where: Q = Heat energy, m = mass, c = specific heat capacity, ΔT = change in temperature
Since the mass is not given, we need to solve for it.
44000 kJ = m * 8.37 J/g-°C * 15°C
Convert 44000 kJ to J: 44,000,000 J
44,000,000 J = m * 8.37 J/g-°C * 15°C
First, let's convert the temperature change to Kelvin.
ΔT = 15°C + 273.15 (Kelvin conversion)
ΔT = 288.15 K
Next, we can solve for the mass:
m = (44,000,000 J) / (8.37 J/g-°C * 288.15 °C)
m ≈ 1815.68 g
Step 3: Calculate the volume of the antifreeze.
Density (d) = mass (m) / volume (v)
1.038 g/mL = 1815.68 g / v
v = (1815.68 g) / (1.038 g/mL)
v ≈ 1748.36 mL
Step 4: Convert the volume to liters.
v = 1748.36 mL / 1000
v ≈ 1.74836 L
Step 5: Calculate how many times the cooling system turns over its volume of antifreeze.
Number of turns = volume of antifreeze / capacity of the cooling system
Number of turns = 1.74836 L / 8.40 L
Number of turns ≈ 0.2082 turns
So, the correct answer is approximately 0.2082 turns, which represents how many times the cooling system turns over its volume of antifreeze to maintain the engine temperature during its hour of operation.