A 10 kg iron bar (specific heat 0.11 cal/g`""^(@)`C) at 80`""^(@)`C is placed on a block of ice. How much ice melts?

Options

A. 1.1 kg
B. 10 kg
C. 16 kg
D. 60 kg

Explanation

To solve the problem of how much ice melts when a 10 kg iron bar at 80°C is placed on it, we will follow these steps:

### Step 1: Calculate the heat lost by the iron bar.
The formula to calculate the heat lost (Q) by the iron bar is given by:

\[ Q = m \cdot c \cdot \Delta T \]

where:
- \( m \) = mass of the iron bar (in grams)
- \( c \) = specific heat of iron (in cal/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of the iron bar = 10 kg = 10,000 g
- Specific heat of iron = 0.11 cal/g°C
- Initial temperature of the iron bar = 80°C
- Final temperature (when it reaches thermal equilibrium with ice) = 0°C

Now, calculate the change in temperature:

\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 80°C - 0°C = 80°C \]

Now substitute the values into the formula:

\[ Q = 10,000 \, \text{g} \cdot 0.11 \, \text{cal/g°C} \cdot 80°C \]

\[ Q = 10,000 \cdot 0.11 \cdot 80 = 88,000 \, \text{cal} \]

### Step 2: Calculate the mass of ice melted using the heat gained by the ice.
The heat gained by the ice (which melts it) can be calculated using the formula:

\[ Q = m_{\text{ice}} \cdot L \]

where:
- \( m_{\text{ice}} \) = mass of ice melted (in grams)
- \( L \) = latent heat of fusion of ice (in cal/g)

The latent heat of fusion of ice is approximately 80 cal/g.

Setting the heat lost by the iron bar equal to the heat gained by the ice:

\[ 88,000 \, \text{cal} = m_{\text{ice}} \cdot 80 \, \text{cal/g} \]

Now, solve for \( m_{\text{ice}} \):

\[ m_{\text{ice}} = \frac{88,000 \, \text{cal}}{80 \, \text{cal/g}} = 1,100 \, \text{g} \]

### Step 3: Convert grams to kilograms.
Since the question asks for the mass of ice melted in kilograms:

\[ m_{\text{ice}} = \frac{1,100 \, \text{g}}{1,000} = 1.1 \, \text{kg} \]

### Final Answer:
The mass of ice that melts is **1.1 kg**.

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