A beaker contains 40 g of water at `20^@C`. Now 50 g of ice is put into the beaker. The resulting temperature will be

Options

A. `-7^@C`
B. `0^@C`
C. `10^@C`
D. `1.5^@C`

Answer

To solve the problem of finding the resulting temperature when 50 g of ice is added to 40 g of water at 20°C, we can follow these steps:

### Step 1: Calculate the heat released by the water
The formula to calculate the heat released by the water is:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q \) = heat released (in calories)
- \( m \) = mass of water (in grams)
- \( c \) = specific heat of water (1 cal/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of water, \( m = 40 \, \text{g} \)
- Initial temperature of water, \( T_i = 20 \, \text{°C} \)
- Final temperature of water, \( T_f = 0 \, \text{°C} \) (since the ice will melt and the system will reach thermal equilibrium at 0°C)

The change in temperature, \( \Delta T = T_i - T_f = 20 - 0 = 20 \, \text{°C} \).

Now, substituting the values into the formula:
\[ Q = 40 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 800 \, \text{cal} \]

### Step 2: Calculate the heat gained by the ice
The heat gained by the ice when it melts is given by:
\[ Q = m \cdot L_f \]
where:
- \( L_f \) = latent heat of fusion of ice (approximately 80 cal/g)

Given:
- Mass of ice, \( m = 50 \, \text{g} \)

Now substituting the values:
\[ Q = 50 \, \text{g} \cdot 80 \, \text{cal/g} = 4000 \, \text{cal} \]

### Step 3: Compare heat released and heat gained
- Heat released by water = 800 cal
- Heat gained by ice = 4000 cal

Since the heat gained by the ice (4000 cal) is greater than the heat released by the water (800 cal), all the heat from the water will be used to melt the ice, and the remaining heat will not be sufficient to raise the temperature of the resulting water above 0°C.

### Conclusion
The resulting temperature of the mixture will be:
\[ \text{Resulting Temperature} = 0 \, \text{°C} \]