The enthalpy of combustion of mol. Wt. 180 glucose is -2808 KJ `"mol"^(-1)` at `25^(@)`C . X and Y grams of glucose do you need to consume respectively cases [Assume wt=62.5 Kg].
(a) to climb a flight of stairs rising through 3M.
to climb a mountain of altitude 3000 M?
Assume that 25% of enthalpy can be converted to useful work.
X and Y are related as X=mY, then find m.

Explanation

To solve the problem, we need to calculate the amount of glucose (in grams) required to climb a flight of stairs (3 meters) and a mountain (3000 meters) using the enthalpy of combustion of glucose.

### Given Data:
- Enthalpy of combustion of glucose = -2808 kJ/mol
- Molecular weight of glucose = 180 g/mol
- Weight of the person = 62.5 kg
- Gravitational acceleration (g) = 9.81 m/s²
- Useful work conversion efficiency = 25%

### Step 1: Calculate the energy required to climb 3 meters.
The energy required (E) to lift a mass (m) to a height (h) is given by the formula:
\[ E = m \cdot g \cdot h \]

For climbing 3 meters:
- m = 62.5 kg
- g = 9.81 m/s²
- h = 3 m

Substituting the values:
\[ E = 62.5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 3 \, \text{m} \]
\[ E = 62.5 \cdot 9.81 \cdot 3 \]
\[ E = 1848.375 \, \text{J} \]
\[ E = 1.848375 \, \text{kJ} \]

### Step 2: Calculate the amount of glucose needed (X) for climbing 3 meters.
Since only 25% of the energy from the combustion of glucose is converted to useful work:
\[ \text{Useful energy from glucose} = 0.25 \cdot \text{Enthalpy of combustion} \]

The energy released from 1 mole of glucose:
\[ \text{Energy from 1 mole of glucose} = 0.25 \cdot (-2808 \, \text{kJ}) = -702 \, \text{kJ} \]

Now, we can find out how many moles of glucose are needed to provide 1.848375 kJ:
Let \( n \) be the number of moles of glucose needed:
\[ n \cdot (-702) = -1.848375 \]
\[ n = \frac{1.848375}{702} \]
\[ n \approx 0.00263 \, \text{moles} \]

Now, convert moles to grams:
\[ \text{Mass of glucose (X)} = n \cdot \text{Molar mass} = 0.00263 \cdot 180 \]
\[ X \approx 0.474 \, \text{grams} \]

### Step 3: Calculate the energy required to climb 3000 meters.
Using the same formula as before:
\[ E = 62.5 \cdot 9.81 \cdot 3000 \]
\[ E = 62.5 \cdot 9.81 \cdot 3000 = 1833750 \, \text{J} \]
\[ E = 1833.75 \, \text{kJ} \]

### Step 4: Calculate the amount of glucose needed (Y) for climbing 3000 meters.
Using the same useful energy conversion:
\[ n \cdot (-702) = -1833.75 \]
\[ n = \frac{1833.75}{702} \]
\[ n \approx 2.61 \, \text{moles} \]

Convert moles to grams:
\[ \text{Mass of glucose (Y)} = n \cdot \text{Molar mass} = 2.61 \cdot 180 \]
\[ Y \approx 471.8 \, \text{grams} \]

### Step 5: Relate X and Y.
Given that \( X = mY \):
\[ 0.474 = m \cdot 471.8 \]
\[ m = \frac{0.474}{471.8} \]
\[ m \approx 0.001 \]

### Final Answers:
- X (grams of glucose for 3m) = 0.474 grams
- Y (grams of glucose for 3000m) = 471.8 grams
- m = 0.001