A thermos contains 150` cm^(3)` of coffee at `85^(@)C`. To cool the coffee, you drop two 11-g ice cubes into the thermos. The ice cubes are initially at `0^(@)C` and melt completely. What is the final temperature of the coffee ? Treat the coffee as if it were water.

Options

A. `59^(@)C`
B. `77^(@)C`
C. `64^(@)C`
D. `73^(@)C`

Answer

To solve the problem, we will use the principle of conservation of energy. The heat lost by the coffee will be equal to the heat gained by the ice cubes as they melt and then warm up to the final temperature.

### Step 1: Calculate the mass of coffee
Given that the volume of coffee is 150 cm³ and the density of water (and coffee, in this case) is approximately 1 g/cm³, we can find the mass of the coffee.

\[ \text{Mass of coffee} = \text{Volume} \times \text{Density} = 150 \, \text{cm}^3 \times 1 \, \text{g/cm}^3 = 150 \, \text{g} \]

### Step 2: Calculate the total mass of ice
We have two ice cubes, each weighing 11 g. Therefore, the total mass of ice is:

\[ \text{Total mass of ice} = 2 \times 11 \, \text{g} = 22 \, \text{g} \]

### Step 3: Calculate the heat lost by the coffee
The coffee cools from 85°C to the final temperature \( T \). The specific heat capacity of water is approximately 1 cal/g°C. The heat lost by the coffee can be calculated as:

\[ Q_{\text{coffee}} = m_{\text{coffee}} \cdot c \cdot \Delta T = 150 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (85 - T) \]

### Step 4: Calculate the heat gained by the ice
The ice first melts and then warms up to the final temperature \( T \). The latent heat of fusion for ice is approximately 80 cal/g. The heat gained by the ice is the sum of the heat required to melt the ice and the heat required to raise the temperature of the melted ice (now water) from 0°C to \( T \):

\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L + m_{\text{ice}} \cdot c \cdot T = 22 \, \text{g} \cdot 80 \, \text{cal/g} + 22 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot T \]

Calculating this gives:

\[ Q_{\text{ice}} = 1760 \, \text{cal} + 22T \]

### Step 5: Set up the energy balance equation
According to the conservation of energy, the heat lost by the coffee is equal to the heat gained by the ice:

\[ 150(85 - T) = 1760 + 22T \]

### Step 6: Solve for \( T \)
Expanding and rearranging the equation gives:

\[ 12750 - 150T = 1760 + 22T \]

Combining like terms:

\[ 12750 - 1760 = 150T + 22T \]
\[ 10990 = 172T \]

Now, solving for \( T \):

\[ T = \frac{10990}{172} \approx 63.89°C \]

### Final Answer
The final temperature of the coffee after the ice has melted and reached thermal equilibrium is approximately **63.89°C**.