A liquid at $30^\circ\text{C}$ is poured very slowly into a Calorimeter that is at temperature of $110^\circ\text{C}$. The boiling temperature of the liquid is $80^\circ\text{C}$. It is found that the first $5\,\text{gm}$ of the liquid completely evaporates. After pouring another $80\,\text{gm}$ of the liquid to its specific heat will be ___$^\circ\text{C}$. [Neglect the heat exchange with surrounding]

Step 1: Identify the initial parameters of the system.

The given parameters are:

$$T_{\text{liquid}} = 30^\circ\text{C}$$ $$T_{\text{calorimeter}} = 110^\circ\text{C}$$ $$T_{\text{boil}} = 80^\circ\text{C}$$

Step 2: Analyze the heating and evaporation of the first $5\text{ g}$ of liquid.

As the liquid is poured slowly, it absorbs heat from the calorimeter and its temperature rises from $30^\circ\text{C}$ to $80^\circ\text{C}$ according to:

$$Q = mc\Delta T$$

Once the liquid reaches its boiling point of $80^\circ\text{C}$, it begins to evaporate. The heat required for this phase change is:

$$Q = mL$$

Since the first $5\text{ g}$ of the liquid completely evaporates, the liquid has reached its boiling point.

Step 3: Determine the temperature after adding more liquid.

When another $80\text{ g}$ of liquid is added, it continues to boil. During a phase change (boiling), the temperature of the liquid remains constant at its boiling point.

$$T = T_{\text{boil}} = 80^\circ\text{C}$$

Answer: $80^\circ\text{C}$