The temperature of 100g of water is to be raised from `24^@C` to `90^@C` by adding steam to it. Calculate the mass of the steam required for this purpose.

Answer

To solve the problem of raising the temperature of 100g of water from \(24^\circ C\) to \(90^\circ C\) by adding steam, we will follow these steps:

### Step 1: Calculate the heat gained by the water
The heat gained by the water (\(Q_w\)) can be calculated using the formula:
\[ Q_w = mc\Delta T \]
where:
- \(m\) is the mass of the water (in kg),
- \(c\) is the specific heat capacity of water (approximately \(4200 \, \text{J/kg} \cdot \text{°C}\)),
- \(\Delta T\) is the change in temperature.

Given:
- Mass of water = \(100 \, \text{g} = 0.1 \, \text{kg}\)
- Initial temperature = \(24^\circ C\)
- Final temperature = \(90^\circ C\)
- Change in temperature (\(\Delta T\)) = \(90 - 24 = 66^\circ C\)

Now substituting the values:
\[ Q_w = 0.1 \, \text{kg} \times 4200 \, \text{J/kg} \cdot \text{°C} \times 66 \, \text{°C} \]
Calculating this gives:
\[ Q_w = 0.1 \times 4200 \times 66 = 27720 \, \text{J} \]

### Step 2: Relate the heat gained by water to the heat lost by steam
When steam condenses to water, it releases heat. The heat lost by the steam (\(Q_s\)) can be expressed as:
\[ Q_s = m_s L \]
where:
- \(m_s\) is the mass of the steam (in kg),
- \(L\) is the latent heat of vaporization of water (approximately \(540 \, \text{kJ/kg} = 540000 \, \text{J/kg}\)).

### Step 3: 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 write:
\[ Q_w = Q_s \]
Substituting the expressions we have:
\[ 27720 \, \text{J} = m_s \times 540000 \, \text{J/kg} \]

### Step 4: Solve for the mass of steam (\(m_s\))
Rearranging the equation to find \(m_s\):
\[ m_s = \frac{27720 \, \text{J}}{540000 \, \text{J/kg}} \]
Calculating this gives:
\[ m_s = \frac{27720}{540000} \approx 0.0514 \, \text{kg} \]

### Step 5: Convert the mass of steam to grams
To express the mass of steam in grams:
\[ m_s \approx 0.0514 \, \text{kg} \times 1000 \, \text{g/kg} = 51.4 \, \text{g} \]

### Final Answer
The mass of steam required to raise the temperature of 100g of water from \(24^\circ C\) to \(90^\circ C\) is approximately **51.4 grams**.

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