What is the resulting temperature when 150 g of ice at `0^(@)C` mixed with 300 g of water at `50^(@)C` ?

Options

A. `25^(@)C`
B. `33.3^(@)C`
C. `13.4^(@)C`
D. `6.6^(@)C `

Answer

To solve the problem of finding the resulting temperature when 150 g of ice at 0°C is mixed with 300 g of water at 50°C, we can follow these steps:

### Step 1: Identify the known values
- Mass of ice (m_ice) = 150 g
- Initial temperature of ice (T_ice) = 0°C
- Mass of water (m_water) = 300 g
- Initial temperature of water (T_water) = 50°C
- Specific heat of water (c_water) = 1 cal/g°C
- Latent heat of fusion of ice (L_f) = 80 cal/g

### Step 2: Set up the heat transfer equation
When ice is added to water, the heat lost by the water will equal the heat gained by the ice to melt and then the heat gained by the resulting water from the melted ice.

Let the final temperature of the mixture be T°C.

The heat lost by the water can be expressed as:
\[ Q_{\text{lost}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial water}} - T) \]
\[ Q_{\text{lost}} = 300 \cdot 1 \cdot (50 - T) \]

The heat gained by the ice can be expressed as:
1. Heat gained to melt the ice:
\[ Q_{\text{melt}} = m_{\text{ice}} \cdot L_f \]
\[ Q_{\text{melt}} = 150 \cdot 80 \]

2. Heat gained by the melted ice (now water) to reach the final temperature T:
\[ Q_{\text{gained}} = m_{\text{ice}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial ice}}) \]
\[ Q_{\text{gained}} = 150 \cdot 1 \cdot (T - 0) \]

### Step 3: Combine the equations
The total heat gained by the ice is:
\[ Q_{\text{gained}} = Q_{\text{melt}} + Q_{\text{gained}} \]
\[ Q_{\text{gained}} = (150 \cdot 80) + (150 \cdot T) \]

Setting the heat lost equal to the heat gained:
\[ 300 \cdot (50 - T) = (150 \cdot 80) + (150 \cdot T) \]

### Step 4: Solve for T
Expanding both sides:
\[ 15000 - 300T = 12000 + 150T \]

Rearranging the equation:
\[ 15000 - 12000 = 300T + 150T \]
\[ 3000 = 450T \]

Dividing both sides by 450:
\[ T = \frac{3000}{450} \]
\[ T = 6.67°C \]

### Final Result
The resulting temperature when 150 g of ice at 0°C is mixed with 300 g of water at 50°C is approximately **6.67°C**.