One end of a copper rod of length 1.0 m and area of cross-section `10^(-3)` is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is `92 cal//m-s-.^(@)C` and the latent heat of ice is `8xx 10^(4) cal//kg`, then the amount of ice which will melt in one minute is

Options

A. `9.2 xx10^(-3)` kg
B. `8 xx10^(-3)` kg
C. `6.9 xx10^(-3)` kg
D. `5.4 xx10^(-3)` kg

Answer

To solve the problem, we will use the formula for heat transfer through conduction and the relationship between heat, mass, and latent heat of fusion.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the copper rod, \( L = 1.0 \, \text{m} \)
- Area of cross-section, \( A = 10^{-3} \, \text{m}^2 \)
- Coefficient of thermal conductivity of copper, \( k = 92 \, \text{cal/(m·s·°C)} \)
- Temperature difference, \( \Delta T = 100°C - 0°C = 100°C \)
- Latent heat of ice, \( L_f = 8 \times 10^4 \, \text{cal/kg} \)
- Time, \( t = 1 \, \text{minute} = 60 \, \text{s} \)

2. **Calculate the heat transferred (Q) through the rod:**
The formula for heat transfer by conduction is given by:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \cdot t \]
Here, \( L \) is the length of the rod.

3. **Substituting the values into the equation:**
\[ Q = \frac{92 \, \text{cal/(m·s·°C)} \cdot 10^{-3} \, \text{m}^2 \cdot 100 \, °C}{1 \, \text{m}} \cdot 60 \, \text{s} \]

4. **Calculating Q:**
\[ Q = 92 \cdot 10^{-3} \cdot 100 \cdot 60 \]
\[ Q = 92 \cdot 10^{-3} \cdot 6000 \]
\[ Q = 92 \cdot 6 = 552 \, \text{cal} \]

5. **Using the heat to find the mass of ice melted (m):**
The heat required to melt ice is given by:
\[ Q = m \cdot L_f \]
Rearranging gives:
\[ m = \frac{Q}{L_f} \]

6. **Substituting the values:**
\[ m = \frac{552 \, \text{cal}}{8 \times 10^4 \, \text{cal/kg}} \]

7. **Calculating m:**
\[ m = \frac{552}{80000} = 0.0069 \, \text{kg} \]

8. **Final answer:**
The amount of ice that will melt in one minute is:
\[ m = 6.9 \times 10^{-3} \, \text{kg} \]