To calculate the amount of heat required to change 1 g of ice at -10°C to steam at 120°C, we will break the process down into several steps, calculating the heat required for each step and then summing them up.
### Step 1: Heating the Ice from -10°C to 0°C
We need to heat the ice from -10°C to 0°C. The specific heat of ice is given as \( C_{ice} = 0.5 \, \text{cal/g°C} \).
\[
Q_1 = m \cdot C_{ice} \cdot \Delta T
\]
\[
Q_1 = 1 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot (0 - (-10)) \, \text{°C}
\]
\[
Q_1 = 1 \cdot 0.5 \cdot 10 = 5 \, \text{cal}
\]
### Step 2: Melting the Ice at 0°C
Next, we need to melt the ice at 0°C. The latent heat of fusion of ice is given as \( L_f = 80 \, \text{cal/g} \).
\[
Q_2 = m \cdot L_f
\]
\[
Q_2 = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal}
\]
### Step 3: Heating the Water from 0°C to 100°C
Now, we need to heat the water from 0°C to 100°C. The specific heat of water is given as \( C_{water} = 1.0 \, \text{cal/g°C} \).
\[
Q_3 = m \cdot C_{water} \cdot \Delta T
\]
\[
Q_3 = 1 \, \text{g} \cdot 1.0 \, \text{cal/g°C} \cdot (100 - 0) \, \text{°C}
\]
\[
Q_3 = 1 \cdot 1.0 \cdot 100 = 100 \, \text{cal}
\]
### Step 4: Vaporizing the Water at 100°C
Next, we need to convert the water at 100°C to steam. The latent heat of vaporization of water is given as \( L_v = 540 \, \text{cal/g} \).
\[
Q_4 = m \cdot L_v
\]
\[
Q_4 = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal}
\]
### Step 5: Heating the Steam from 100°C to 120°C
Finally, we need to heat the steam from 100°C to 120°C. We first calculate the molar specific heat of steam, which is given as \( C_{p, steam} = 4R \) where \( R = 2 \, \text{cal/mol K} \).
\[
C_{p, steam} = 4 \cdot 2 = 8 \, \text{cal/mol K}
\]
Since the molar mass of water is 18 g/mol, the specific heat of steam in cal/g°C is:
\[
C_{p, steam} = \frac{8 \, \text{cal/mol K}}{18 \, \text{g/mol}} \approx 0.44 \, \text{cal/g°C}
\]
Now we can calculate the heat required to heat the steam from 100°C to 120°C:
\[
Q_5 = m \cdot C_{p, steam} \cdot \Delta T
\]
\[
Q_5 = 1 \, \text{g} \cdot 0.44 \, \text{cal/g°C} \cdot (120 - 100) \, \text{°C}
\]
\[
Q_5 = 1 \cdot 0.44 \cdot 20 = 8.8 \, \text{cal}
\]
### Total Heat Required
Now, we can sum all the heat quantities calculated in each step:
\[
Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5
\]
\[
Q_{total} = 5 + 80 + 100 + 540 + 8.8 = 733.8 \, \text{cal}
\]
### Final Answer
The total amount of heat required to change 1 g of ice at -10°C to steam at 120°C is **733.8 calories**.
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