2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg per °C and 0.5 kcal/kg/°C while the latent heat of fusion of ice is 80 kcal/kg

Correct Answer: The correct option is 6 kg.

To find the final mass of water remaining in the insulating vessel, we can apply the principle of calorimetry (heat lost = heat gained) to analyze the heat exchange between the ice and the water.

1. Identify the given parameters:
Mass of ice,
$m_{\text{ice}}=2\text{kg}$
Initial temperature of ice,
$T_{\text{ice}}=-20\text{°C}$
Specific heat of ice,
$c_{\text{ice}}=0.5\text{kcal/kg°C}$
Mass of water,
$m_{\text{water}}=5\text{kg}$
Initial temperature of water,
$T_{\text{water}}=20\text{°C}$
Specific heat of water,
$c_{\text{water}}=1.0\text{kcal/kg°C}$
Latent heat of fusion of ice,
$L=80\text{kcal/kg}$

2. Calculate the heat required to warm the ice:
First, let's find the heat required to raise the temperature of the ice from $-20\text{°C}$ to $0\text{°C}$:
$Q_1=m_{\text{ice}}\cdot c_{\text{ice}}\cdot \Delta T$
$Q_1=2\cdot 0.5\cdot (0-(-20\left)\right)$
$Q_1=2\cdot 0.5\cdot 20=20\text{kcal}$

3. Calculate the maximum heat that the water can release:
Now, let's find the heat released by the water if it cools down from $20\text{°C}$ to $0\text{°C}$:
$Q_2=m_{\text{water}}\cdot c_{\text{water}}\cdot \Delta T$
$Q_2=5\cdot 1.0\cdot (20-0)$
$Q_2=5\cdot 1.0\cdot 20=100\text{kcal}$

4. Determine the phase change of the ice:
Since the maximum heat available from cooling the water to $0\text{°C}$ (100 kcal) is greater than the heat required to warm the ice to $0\text{°C}$ (20 kcal), all the ice will reach $0\text{°C}$ , and the remaining heat will melt a portion of the ice.
The excess heat available for melting the ice is:
$Q_{\text{excess}}=Q_2-Q_1$
$Q_{\text{excess}}=100-20=80\text{kcal}$

Let $m$ be the mass of ice that melts. The heat required to melt this mass is:
$Q_{\text{melt}}=m\cdot L$
Setting $Q_{\text{melt}}=Q_{\text{excess}}$:
$m\cdot 80=80$
$m=1\text{kg}$

Thus, 1 kg of ice melts into water at $0\text{°C}$ . The mixture reaches a state of thermal equilibrium at $0\text{°C}$ with both ice and water present.

5. Calculate the final mass of water:
The final mass of water in the container is the sum of the initial water mass and the mass of the melted ice:
$m_{\text{final water}}=m_{\text{water}}+m$
$m_{\text{final water}}=5\text{kg}+1\text{kg}=6\text{kg}$