How much heat is required to convert 2g of ice at 0°C into steam at 100°C

Options

A. 860 cal
B. 1440 cal
C. 320 cal
D. 160 cal

Explanation

To solve the problem of how much heat is required to convert 2g of ice at 0°C into steam at 100°C, we need to break down the process into several steps. Each step involves a phase change or a temperature change, and we will use the specific heat capacities and latent heats for the calculations.

### Step-by-Step Solution:

1. **Melting Ice to Water at 0°C**:
- We need to convert ice at 0°C to water at 0°C.
- The latent heat of fusion (L_f) for ice is approximately 334 J/g.
- Heat required (Q1) = mass × L_f = 2g × 334 J/g = 668 J.

2. **Heating Water from 0°C to 100°C**:
- Next, we need to heat the water from 0°C to 100°C.
- The specific heat capacity (c) of water is approximately 4.18 J/g°C.
- Temperature change (ΔT) = 100°C - 0°C = 100°C.
- Heat required (Q2) = mass × c × ΔT = 2g × 4.18 J/g°C × 100°C = 836 J.

3. **Converting Water to Steam at 100°C**:
- Now, we convert the water at 100°C to steam at 100°C.
- The latent heat of vaporization (L_v) for water is approximately 2260 J/g.
- Heat required (Q3) = mass × L_v = 2g × 2260 J/g = 4520 J.

4. **Total Heat Required**:
- The total heat required (Q_total) is the sum of all the heat quantities calculated in the previous steps:
\[ Q_{total} = Q_1 + Q_2 + Q_3 = 668 J + 836 J + 4520 J = 6024 J. \]

### Final Answer:
The total heat required to convert 2g of ice at 0°C into steam at 100°C is **6024 Joules**.