To find the total amount of heat required to convert 10 g of ice at -10°C into steam at 100°C, we will break the process down into several steps:
### Step 1: Heating the Ice from -10°C to 0°C
We need to calculate the heat required to raise the temperature of the ice from -10°C to 0°C. The formula for heat (Q) is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
Where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( c = 0.5 \, \text{cal g}^{-1} \text{°C}^{-1} \) (specific heat capacity of ice)
- \( \Delta T = 0 - (-10) = 10 \, \text{°C} \) (change in temperature)
Calculating \( Q_1 \):
\[ Q_1 = 10 \, \text{g} \cdot 0.5 \, \text{cal g}^{-1} \text{°C}^{-1} \cdot 10 \, \text{°C} \]
\[ Q_1 = 10 \cdot 0.5 \cdot 10 = 50 \, \text{cal} \]
### Step 2: Melting the Ice at 0°C
Next, we convert the ice at 0°C to water at 0°C using the latent heat of fusion:
\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f = 80 \, \text{cal g}^{-1} \) (latent heat of fusion)
Calculating \( Q_2 \):
\[ Q_2 = 10 \, \text{g} \cdot 80 \, \text{cal g}^{-1} \]
\[ Q_2 = 800 \, \text{cal} \]
### Step 3: Heating the Water from 0°C to 100°C
Now, we need to heat the water from 0°C to 100°C:
\[ Q_3 = m \cdot c \cdot \Delta T \]
Where:
- \( c = 1 \, \text{cal g}^{-1} \text{°C}^{-1} \) (specific heat capacity of water)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \)
Calculating \( Q_3 \):
\[ Q_3 = 10 \, \text{g} \cdot 1 \, \text{cal g}^{-1} \text{°C}^{-1} \cdot 100 \, \text{°C} \]
\[ Q_3 = 1000 \, \text{cal} \]
### Step 4: Vaporizing the Water at 100°C
Finally, we convert the water at 100°C to steam at 100°C using the latent heat of vaporization:
\[ Q_4 = m \cdot L_v \]
Where:
- \( L_v = 540 \, \text{cal g}^{-1} \) (latent heat of vaporization)
Calculating \( Q_4 \):
\[ Q_4 = 10 \, \text{g} \cdot 540 \, \text{cal g}^{-1} \]
\[ Q_4 = 5400 \, \text{cal} \]
### Step 5: Total Heat Required
Now, we can find the total heat required by summing all the heat quantities calculated:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 \]
\[ Q_{\text{total}} = 50 \, \text{cal} + 800 \, \text{cal} + 1000 \, \text{cal} + 5400 \, \text{cal} \]
\[ Q_{\text{total}} = 7250 \, \text{cal} \]
Thus, the total amount of heat required to convert 10 g of ice at -10°C into steam at 100°C is **7250 calories**.
