The amount of heat required to raise the temperature of 75 kg of ice at `0^oC` to water at `10^oC` is (latent heat of fusion of ice is 80 cal/g, specific heat of water is 1 cal/`g^oC`)

Options

A. 750 kcal
B. `6.75xx10^3kcal`
C. `60xx10^3kcal`
D. 6.75 kcal

Answer

To solve the problem of finding the amount of heat required to raise the temperature of 75 kg of ice at 0°C to water at 10°C, we need to consider two main steps:

1. **Melting the Ice**: We first need to convert the ice at 0°C to water at 0°C.
2. **Heating the Water**: Next, we will heat the resulting water from 0°C to 10°C.

### Step 1: Calculate the Heat Required to Melt the Ice

The formula for calculating the heat required to melt ice is given by:

\[ Q_1 = m \cdot L_f \]

Where:
- \( Q_1 \) = heat required to melt the ice (in calories)
- \( m \) = mass of ice (in grams)
- \( L_f \) = latent heat of fusion of ice (in cal/g)

Given:
- Mass of ice = 75 kg = 75,000 g (since 1 kg = 1000 g)
- Latent heat of fusion of ice \( L_f = 80 \, \text{cal/g} \)

Now substituting the values:

\[ Q_1 = 75000 \, \text{g} \cdot 80 \, \text{cal/g} \]
\[ Q_1 = 6000000 \, \text{cal} \]

### Step 2: Calculate the Heat Required to Raise the Temperature of Water

The formula for calculating the heat required to raise the temperature of water is given by:

\[ Q_2 = m \cdot c \cdot \Delta T \]

Where:
- \( Q_2 \) = heat required to raise the temperature (in calories)
- \( m \) = mass of water (in grams)
- \( c \) = specific heat of water (in cal/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of water = 75,000 g (since the mass of ice and water remains the same)
- Specific heat of water \( c = 1 \, \text{cal/g°C} \)
- Change in temperature \( \Delta T = 10°C - 0°C = 10°C \)

Now substituting the values:

\[ Q_2 = 75000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 10°C \]
\[ Q_2 = 750000 \, \text{cal} \]

### Step 3: Total Heat Required

Now, we can find the total heat required \( Q \) by adding \( Q_1 \) and \( Q_2 \):

\[ Q = Q_1 + Q_2 \]
\[ Q = 6000000 \, \text{cal} + 750000 \, \text{cal} \]
\[ Q = 6750000 \, \text{cal} \]

### Final Answer

The total amount of heat required to raise the temperature of 75 kg of ice at 0°C to water at 10°C is **6,750,000 calories**.

---