Steam at `100^(@)C` is added slowly to 1400 gm of water at `16^(@)C` until the temperature of water is raised to `80^(@)C`. The mass of steam required to do this is (`L_(V)`= 540 cal/g)

Options

A. A) 160 g
B. B) 120 g
C. C) 250 g
D. D) 320 g

Answer

To solve the problem, we need to find the mass of steam required to raise the temperature of 1400 grams of water from 16°C to 80°C. We will use the principle of conservation of energy, where the heat gained by the water is equal to the heat lost by the steam.

### Step-by-Step Solution:

1. **Calculate the heat gained by the water (Q1)**:
The formula for heat gained or lost is given by:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of the water (1400 g)
- \( c \) = specific heat capacity of water (1 cal/g°C)
- \( \Delta T \) = change in temperature = final temperature - initial temperature = \( 80°C - 16°C = 64°C \)

Substituting the values:
\[ Q_1 = 1400 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 64 \, \text{°C} = 89600 \, \text{cal} \]

2. **Calculate the heat lost by the steam (Q2)**:
When steam condenses, it releases heat. The heat released by the steam can be calculated using:
\[ Q_2 = m \cdot L_V + m \cdot c \cdot (T_{steam} - T_{final}) \]
where:
- \( m \) = mass of the steam (unknown)
- \( L_V \) = latent heat of vaporization of steam (540 cal/g)
- \( T_{steam} \) = temperature of steam (100°C)
- \( T_{final} \) = final temperature of water (80°C)

Therefore, the equation becomes:
\[ Q_2 = m \cdot 540 \, \text{cal/g} + m \cdot 1 \, \text{cal/g°C} \cdot (100°C - 80°C) \]
Simplifying:
\[ Q_2 = m \cdot 540 + m \cdot 20 = m \cdot (540 + 20) = m \cdot 560 \, \text{cal} \]

3. **Set the heat gained equal to the heat lost**:
According to the conservation of energy:
\[ Q_1 = Q_2 \]
Thus:
\[ 89600 \, \text{cal} = m \cdot 560 \, \text{cal} \]

4. **Solve for the mass of the steam (m)**:
Rearranging the equation gives:
\[ m = \frac{89600 \, \text{cal}}{560 \, \text{cal/g}} = 160 \, \text{g} \]

### Final Answer:
The mass of steam required to raise the temperature of the water to 80°C is **160 grams**.

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