How many grams of carbohydrate does a person of mass 75 kg need to metabolize to climb five flights of stairs (15 m height increase)? Each gram of carbohydrate provides 17.6 kJ of energy. Assume 10.0% efficiency—that is, 10.0% of the available chemical energy in the carbohydrate is converted to mechanical energy.
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What happens to the other 90% of the energy?
Could someone show me how to do this please. Thanks. I have a hard time figuring it out.
The person needs to do an amount of work of W = 75 kg * 15 m * 9.81 m/s^2 against gravity.
Per gram of carbohydrates, you have available 10% of 17.6 KJ to do mechanical work, so you need to divide W by 1.76 KJ to find the amount of carbohydrates needed in grams.
The other 90% converted to heat.
To calculate the grams of carbohydrate needed to climb five flights of stairs, we need to determine the total energy required for this activity and then account for the 10% efficiency. Here's the step-by-step calculation:
Step 1: Calculate the total energy required
To climb a 15 m height increase, we need to overcome the potential energy difference. The potential energy (PE) is given by the formula PE = mgh, where m is the mass, g is the gravitational acceleration (approximately 9.8 m/s^2), and h is the height. Let's calculate the potential energy:
PE = mgh
PE = (75 kg) * (9.8 m/s^2) * (15 m)
PE = 11,025 J (joules)
Step 2: Convert energy to kilojoules (kJ)
The potential energy is currently in joules, but since we are given the energy content of carbohydrate in kJ, we need to convert joules to kJ. Since 1 kJ = 1000 J:
Potential energy in kJ = 11,025 J / 1000
Potential energy in kJ = 11.025 kJ
Step 3: Calculate the total energy intake needed
Given that 10% of the energy is converted to mechanical energy, the remaining 90% is not used for climbing stairs. Therefore:
Total energy intake needed = Potential energy / Efficiency
Total energy intake needed = 11.025 kJ / 0.1
Total energy intake needed = 110.25 kJ (kilojoules)
Step 4: Convert energy to grams of carbohydrate
We are given that each gram of carbohydrate provides 17.6 kJ of energy. To find the required grams of carbohydrate, divide the total energy intake by the energy content per gram:
Grams of carbohydrate = Total energy intake / Energy content per gram
Grams of carbohydrate = 110.25 kJ / 17.6 kJ/g
Grams of carbohydrate ≈ 6.26 g
Therefore, a person weighing 75 kg needs to metabolize approximately 6.26 grams of carbohydrate to climb five flights of stairs, considering a 10.0% efficiency.
Regarding the fate of the remaining 90% of the energy, it is not used for mechanical work (climbing the stairs) and is converted to other forms, such as heat, sound, or not fully utilized energy in the body.
To calculate the amount of grams of carbohydrate needed to climb the five flights of stairs, we need to consider the energy requirements and efficiency.
First, let's calculate the total energy required to climb the stairs:
Energy (in joules) = Mass (in kg) x Gravitational force (in m/s^2) x Height (in meters)
Energy = 75 kg x 9.8 m/s^2 x 15 m
Energy = 11025 joules
Since each gram of carbohydrate provides 17.6 kJ (kilojoules) of energy, we need to convert the energy to kilojoules:
Energy = 11025 joules = 11.025 kJ
Now, let's account for the efficiency of energy conversion. We are given that the efficiency is 10.0%, meaning only 10.0% of the available chemical energy in the carbohydrate is converted to mechanical energy. The rest is lost as heat and metabolic processes.
So, we need to calculate the amount of carbohydrate needed to provide the required energy:
Carbohydrate (in grams) = Energy (in kJ) / Energy provided per gram (in kJ/g)
Carbohydrate = 11.025 kJ / 17.6 kJ/g
Carbohydrate = 0.625 grams
Therefore, a person of mass 75 kg would need approximately 0.625 grams of carbohydrate to metabolize in order to climb five flights of stairs.
As for the question about the other 90% of the energy, it is essentially lost through heat production and other metabolic processes within the body. Energy conversions are typically not 100% efficient due to various factors, such as friction and inefficiencies in biochemical reactions.