A 52 g mountain climber, start from rest, climbs a vertical distance of 730 m. At top, she is again at rest. In the process, her body generates 4.1*10^6 J of energy via metabolic processes. Her body acts like a heat engine with efficiency ( W/Q_h.) W is the magnitude of work done by her, and Q_h is magnitude of the input heat. So.. can you tell me her effiency as a heat engine?
They tell you the formula to use for the efficiency:
W/Q_h
They also tell you that
Q_h = 4.1*10^6 J
The formula for work done (W) is M g H, where H = 730 m
You only have to do the indicated calculation. I am quite sure the climber's weight is 52 kg, not 52 g.
I get 9.1% for the efficiency
wow. that was simple. thank you.. i should read the question more carefully next time :D
Well, let me calculate that for you while juggling some numbers!
First, we need to find the work done by the mountain climber. We can use the equation W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height:
W = (0.052 kg)(9.8 m/s^2)(730 m) ≈ 370.672 J
Now, we can calculate the efficiency using the formula (W/Q_h):
Efficiency = (370.672 J) / (4.1 × 10^6 J)
*Clown Bot pulls out a giant calculator with confetti shooting out of the buttons*
Efficiency = 0.000090167
So, after all that number juggling, her efficiency as a heat engine is approximately 0.000090167. Not the most impressive efficiency, but hey, at least she got to enjoy the climb!
To find the efficiency of the mountain climber as a heat engine, we can use the formula:
Efficiency = (Work done) / (Input heat)
We are given the magnitude of work done by the climber as 4.1 * 10^6 J.
To find the input heat, we need to consider the energy generated via metabolic processes. Since energy cannot be created or destroyed, the input heat can be considered equal to the total energy generated. Thus, the input heat is also 4.1 * 10^6 J.
Now, we can calculate the efficiency:
Efficiency = (4.1 * 10^6 J) / (4.1 * 10^6 J)
Efficiency = 1
Therefore, the efficiency of the mountain climber as a heat engine is 1, or 100%.
To find the efficiency of the mountain climber as a heat engine, we can use the formula:
Efficiency (η) = W / Q_h
where η is the efficiency, W is the work done, and Q_h is the input heat.
The work done by the mountain climber can be calculated using the formula:
W = mgh
where m is the mass of the climber, g is the acceleration due to gravity, and h is the height climbed.
Given:
m = 52 g = 0.052 kg
g = 9.8 m/s²
h = 730 m
Substituting these values into the equation, we can calculate the work done:
W = (0.052 kg) * (9.8 m/s²) * (730 m)
W = 377.104 J
The input heat (Q_h) is given as 4.1 × 10^6 J.
Now, we can calculate the efficiency:
η = W / Q_h
η = 377.104 J / (4.1 × 10^6 J)
To simplify this calculation, let's express 4.1 × 10^6 J as scientific notation:
η = 377.104 J / (4.1 × 10^6 J)
η = 377.104 J / (4.1 × 10^6 J)
η = 377.104 / 4.1 × 10^6
η ≈ 0.000092 or 9.2 × 10^(-5)
Therefore, the efficiency of the mountain climber as a heat engine is approximately 9.2 × 10^(-5) or 0.0092%.