In an industrial process 10 kg of water per hour is to be heated from `20^@C` to `80^@C`. To do this steam at `150^@C` is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at `90^@C`. How many kilograms of steam is required per hour (specific heat of steam`=1 cal//g^@C`, Latent heat of vapourization`=540 cal//g`)?

Options

A. 1 g
B. 1 kg
C. 10 g
D. 10 kg

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat required to heat the water
We need to calculate the heat required to raise the temperature of 10 kg of water from 20°C to 80°C.

The formula to calculate heat (Q) is:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of water = 10 kg = 10,000 g (since 1 kg = 1000 g)
- \( c \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = \( 80°C - 20°C = 60°C \)

Substituting the values:
\[ Q = 10,000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 60°C \]
\[ Q = 600,000 \, \text{cal} \]

### Step 2: Calculate the heat released by the steam
Next, we need to calculate the heat released by the steam as it condenses and cools down.

The heat released by the steam can be calculated in two parts:
1. Heat released during condensation from 150°C to 100°C.
2. Heat released while cooling from 100°C to 90°C.

#### Part 1: Heat released during condensation
The heat released during condensation (Q1) can be calculated as:
\[ Q_1 = m_s \cdot L_v \]
Where:
- \( m_s \) = mass of steam (in grams)
- \( L_v \) = latent heat of vaporization = 540 cal/g

#### Part 2: Heat released while cooling
The heat released while cooling from 100°C to 90°C (Q2) can be calculated as:
\[ Q_2 = m_s \cdot c \cdot \Delta T \]
Where:
- \( \Delta T \) = change in temperature = \( 100°C - 90°C = 10°C \)

So:
\[ Q_2 = m_s \cdot 1 \, \text{cal/g°C} \cdot 10°C \]
\[ Q_2 = 10 \cdot m_s \]

### Step 3: Total heat released by steam
The total heat released by the steam is:
\[ Q_{\text{total}} = Q_1 + Q_2 \]
\[ Q_{\text{total}} = m_s \cdot 540 + 10 \cdot m_s \]
\[ Q_{\text{total}} = m_s \cdot (540 + 10) \]
\[ Q_{\text{total}} = m_s \cdot 550 \]

### Step 4: Set the heat gained by water equal to the heat lost by steam
Since the heat gained by the water is equal to the heat lost by the steam, we can set the equations equal to each other:
\[ 600,000 = m_s \cdot 550 \]

### Step 5: Solve for the mass of steam
Now, we can solve for \( m_s \):
\[ m_s = \frac{600,000}{550} \]
\[ m_s \approx 1090.91 \, \text{g} \]
Converting grams to kilograms:
\[ m_s \approx 1.09 \, \text{kg} \]

### Conclusion
The mass of steam required per hour is approximately **1.09 kg**.

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