One end of a copper rod of uniform cross section and length 1.5 m is kept in contact with ice and the other end with water at `100^@C`. At what point along its length should a temperature of `200^@C` be maintained so that in the steady state, the mass of ice melting be equal to that of the steam produced in same interval of time. Assume that the whole system is insulated from surroundings:
`[L_("ice")=80 cal//g,L_("steam")=540 cal//g]`

Explanation

To solve the problem step by step, we will analyze the heat transfer through the copper rod and set up equations based on the heat absorbed by ice and the heat released by water.

### Step 1: Understand the setup
We have a copper rod of length 1.5 m with one end at 0°C (ice) and the other end at 100°C (water). We need to maintain a temperature of 200°C at some point along the rod such that the mass of ice melted equals the mass of steam produced.

### Step 2: Define variables
Let:
- \( x \) = distance from the 100°C end to the 200°C point.
- \( L = 1.5 \, \text{m} = 150 \, \text{cm} \) (length of the rod).
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for ice).
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization for steam).

The length from the ice to the 200°C point is \( 1.5 - x \).

### Step 3: Calculate heat transfer to ice
The heat \( Q_1 \) transferred to the ice can be expressed as:
\[ Q_1 = \frac{K A (200 - 0) T}{1.5 - x} \]
This heat is used to melt the ice:
\[ Q_1 = m_1 L_f \]
where \( m_1 \) is the mass of ice melted.

### Step 4: Calculate heat transfer to water
The heat \( Q_2 \) transferred to the water can be expressed as:
\[ Q_2 = \frac{K A (200 - 100) T}{x} \]
This heat is used to convert water to steam:
\[ Q_2 = m_2 L_v \]
where \( m_2 \) is the mass of steam produced.

### Step 5: Equate the masses
Since the mass of ice melted equals the mass of steam produced, we have:
\[ m_1 = m_2 \]

### Step 6: Substitute and simplify
From the equations for \( Q_1 \) and \( Q_2 \):
1. \( m_1 = \frac{K A (200) T}{L_f (1.5 - x)} \)
2. \( m_2 = \frac{K A (100) T}{L_v x} \)

Setting \( m_1 = m_2 \):
\[ \frac{K A (200) T}{80 (1.5 - x)} = \frac{K A (100) T}{540 x} \]
Canceling \( K A T \) from both sides:
\[ \frac{200}{80 (1.5 - x)} = \frac{100}{540 x} \]

### Step 7: Cross-multiply and solve for \( x \)
Cross-multiplying gives:
\[ 200 \cdot 540 x = 100 \cdot 80 (1.5 - x) \]
Expanding both sides:
\[ 108000 x = 8000 (1.5 - x) \]
\[ 108000 x = 12000 - 8000 x \]
Combining like terms:
\[ 108000 x + 8000 x = 12000 \]
\[ 116000 x = 12000 \]
\[ x = \frac{12000}{116000} = \frac{12}{116} = 0.1034 \, \text{m} = 10.34 \, \text{cm} \]

### Step 8: Conclusion
The distance from the 100°C end to the point where the temperature is maintained at 200°C is approximately **10.34 cm**.

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