Ten grams of ice at - `20^(@)C` is to be changed to steam at `130^(@)C` . The specific heat of both ice and steam is 0.5 cal `// ( g^(@)C)`. The specific heat of water is 1.00 cal `// ( g K )`. The heat of fusion is 80 cal `//` g and the heat of vaporization is 540 cal `//` g. The entire process requires.

Options

A. 750cal
B. 1250cal
C. 6950 cal
D. 7450 cal

Answer

To solve the problem of converting 10 grams of ice at -20°C to steam at 130°C, we will break down the process into several steps, calculating the heat required for each step.

### Step 1: Heating the Ice from -20°C to 0°C
We need to calculate the heat required to raise the temperature of ice from -20°C to 0°C.

**Formula:**
\[ Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T \]

Where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( c_{\text{ice}} = 0.5 \, \text{cal/g°C} \) (specific heat of ice)
- \( \Delta T = 0 - (-20) = 20 \, \text{°C} \) (temperature change)

**Calculation:**
\[ Q_1 = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 100 \, \text{cal} \]

### Step 2: Melting the Ice at 0°C to Water
Next, we need to calculate the heat required to melt the ice at 0°C into water.

**Formula:**
\[ Q_2 = m \cdot L_f \]

Where:
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

**Calculation:**
\[ Q_2 = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 3: Heating the Water from 0°C to 100°C
Now, we will calculate the heat required to raise the temperature of water from 0°C to 100°C.

**Formula:**
\[ Q_3 = m \cdot c_{\text{water}} \cdot \Delta T \]

Where:
- \( c_{\text{water}} = 1.0 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \)

**Calculation:**
\[ Q_3 = 10 \, \text{g} \cdot 1.0 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 1000 \, \text{cal} \]

### Step 4: Vaporizing the Water at 100°C to Steam
Next, we will calculate the heat required to convert water at 100°C to steam.

**Formula:**
\[ Q_4 = m \cdot L_v \]

Where:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization)

**Calculation:**
\[ Q_4 = 10 \, \text{g} \cdot 540 \, \text{cal/g} = 5400 \, \text{cal} \]

### Step 5: Heating the Steam from 100°C to 130°C
Finally, we will calculate the heat required to raise the temperature of steam from 100°C to 130°C.

**Formula:**
\[ Q_5 = m \cdot c_{\text{steam}} \cdot \Delta T \]

Where:
- \( c_{\text{steam}} = 0.5 \, \text{cal/g°C} \) (specific heat of steam)
- \( \Delta T = 130 - 100 = 30 \, \text{°C} \)

**Calculation:**
\[ Q_5 = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 30 \, \text{°C} = 150 \, \text{cal} \]

### Total Heat Required
Now, we will sum all the heat values calculated in each step:

\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 \]
\[ Q_{\text{total}} = 100 + 800 + 1000 + 5400 + 150 \]
\[ Q_{\text{total}} = 7450 \, \text{cal} \]

### Final Answer
The total heat required for the entire process is **7450 calories**.

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