Find the volume of iron required to make an open box whose external dimensions are `36cmxx25cmxx16.5cm`, the box beign 1.5 cm thick throughout. If `1 cm^(3)` of iron weighs 8.5 grams, find the weight of the empty box in kilograms.

Answer

To solve the problem of finding the volume of iron required to make an open box and its weight in kilograms, we will follow these steps:

### Step 1: Calculate the Volume of the Outer Box
The external dimensions of the box are given as:
- Length (L) = 36 cm
- Breadth (B) = 25 cm
- Height (H) = 16.5 cm

The volume of the outer box (V_outer) can be calculated using the formula:
\[ V_{\text{outer}} = L \times B \times H \]

Substituting the values:
\[ V_{\text{outer}} = 36 \, \text{cm} \times 25 \, \text{cm} \times 16.5 \, \text{cm} \]
\[ V_{\text{outer}} = 14850 \, \text{cm}^3 \]

### Step 2: Calculate the Dimensions of the Inner Box
The box has a thickness of 1.5 cm throughout. Since it is an open box, we only subtract the thickness from the bottom and the sides.

- Length of the inner box (L_inner):
\[ L_{\text{inner}} = L - 2 \times \text{thickness} = 36 \, \text{cm} - 2 \times 1.5 \, \text{cm} = 36 \, \text{cm} - 3 \, \text{cm} = 33 \, \text{cm} \]

- Breadth of the inner box (B_inner):
\[ B_{\text{inner}} = B - 2 \times \text{thickness} = 25 \, \text{cm} - 2 \times 1.5 \, \text{cm} = 25 \, \text{cm} - 3 \, \text{cm} = 22 \, \text{cm} \]

- Height of the inner box (H_inner):
\[ H_{\text{inner}} = H - \text{thickness} = 16.5 \, \text{cm} - 1.5 \, \text{cm} = 15 \, \text{cm} \]

### Step 3: Calculate the Volume of the Inner Box
Now we can calculate the volume of the inner box (V_inner):
\[ V_{\text{inner}} = L_{\text{inner}} \times B_{\text{inner}} \times H_{\text{inner}} \]
\[ V_{\text{inner}} = 33 \, \text{cm} \times 22 \, \text{cm} \times 15 \, \text{cm} \]
\[ V_{\text{inner}} = 10980 \, \text{cm}^3 \]

### Step 4: Calculate the Volume of Iron Used
The volume of iron (V_iron) required to make the box is the difference between the volume of the outer box and the volume of the inner box:
\[ V_{\text{iron}} = V_{\text{outer}} - V_{\text{inner}} \]
\[ V_{\text{iron}} = 14850 \, \text{cm}^3 - 10980 \, \text{cm}^3 \]
\[ V_{\text{iron}} = 3960 \, \text{cm}^3 \]

### Step 5: Calculate the Weight of the Iron
Given that 1 cm³ of iron weighs 8.5 grams, we can calculate the total weight of the iron:
\[ \text{Weight of iron} = V_{\text{iron}} \times \text{weight per cm}^3 \]
\[ \text{Weight of iron} = 3960 \, \text{cm}^3 \times 8.5 \, \text{grams/cm}^3 \]
\[ \text{Weight of iron} = 33660 \, \text{grams} \]

### Step 6: Convert Weight to Kilograms
To convert grams to kilograms, we divide by 1000:
\[ \text{Weight in kg} = \frac{33660 \, \text{grams}}{1000} \]
\[ \text{Weight in kg} = 33.66 \, \text{kg} \]

### Final Answer
The weight of the empty box is **33.66 kg**.

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