Steam is passed into `54 gm` of water at `30^(@)C` till the temperature of mixture becomes `90^(@)C`. If the latent heat of steam is `536 cal//gm`, the mass of the mixture will be

Options

A. 80 g
B. 60 g
C. 50 g
D. 24 g

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the heat required to raise the temperature of water from 30°C to 90°C.

The formula for heat (Q) required to change the temperature of a substance is given by:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of the substance (in grams)
- \( c \) = specific heat capacity of water (approximately \( 1 \, \text{cal/g°C} \))
- \( \Delta T \) = change in temperature (in °C)

For our case:
- \( m = 54 \, \text{g} \)
- \( c = 1 \, \text{cal/g°C} \)
- \( \Delta T = 90°C - 30°C = 60°C \)

Calculating the heat required:
\[ Q = 54 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 60°C \]
\[ Q = 54 \cdot 60 = 3240 \, \text{cal} \]

### Step 2: Set up the equation for the heat gained by steam.

Let \( m \) be the mass of steam that condenses. The heat gained by the steam when it condenses and cools down to 90°C is given by:
\[ Q = m \cdot L + m \cdot c \cdot (100°C - 90°C) \]
Where:
- \( L = 536 \, \text{cal/g} \) (latent heat of steam)
- \( c = 1 \, \text{cal/g°C} \)

So, the equation becomes:
\[ Q = m \cdot 536 + m \cdot 1 \cdot (100 - 90) \]
\[ Q = m \cdot 536 + m \cdot 10 \]
\[ Q = m \cdot (536 + 10) \]
\[ Q = m \cdot 546 \]

### Step 3: Equate the heat gained by steam to the heat lost by water.

From Step 1, we have:
\[ 3240 = m \cdot 546 \]

### Step 4: Solve for \( m \).

Rearranging the equation gives:
\[ m = \frac{3240}{546} \]

Calculating \( m \):
\[ m \approx 5.93 \, \text{g} \]

### Step 5: Calculate the total mass of the mixture.

The total mass of the mixture is the sum of the mass of water and the mass of steam:
\[ \text{Total mass} = 54 \, \text{g} + 5.93 \, \text{g} \]
\[ \text{Total mass} \approx 59.93 \, \text{g} \]

Rounding this to the nearest whole number gives:
\[ \text{Total mass} \approx 60 \, \text{g} \]

### Final Answer:
The mass of the mixture is approximately **60 grams**.

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