1. Let the calorimeter heat capacity be $C$ (in cal/^ircC), the liquid specific heat be $s$ (in cal g^{-1}ircC$^{-1}$), and latent heat be $L$ (in cal g$^{-1}$).

2. First 5 g of liquid completely evaporates

• Initial temperature of calorimeter = 110^ircC
• Liquid is poured at 30^ircC
• Since it evaporates, it must first be heated from 30^ircC to its boiling point 80^ircC, then vaporized.

So, heat absorbed by 5 g liquid: $Q_1 = 5s(80-30) + 5L = 5(50s + L)$

This heat comes from the calorimeter cooling from 110^ircC to 80^ircC: $Q_1 = C(110-80)=30C$

Hence, $30C = 5(50s+L)$ $6C = 50s + L \quad \text{(1)}$

3. Then another 80 g of liquid is poured

At this stage, the calorimeter is at 80^ircC. The new 80 g liquid is at 30^ircC, and final equilibrium temperature is 50^ircC.

So, heat lost by calorimeter: $Q_2 = C(80-50)=30C$

Heat gained by 80 g liquid: $Q_2 = 80s(50-30)=1600s$

Thus, $30C = 1600s$ $C = \frac{1600s}{30} = \frac{160}{3}s$

4. Substitute into equation (1)

From (1): $6C = 50s + L$ $6\left(\frac{160}{3}s\right)=50s+L$ $320s = 50s + L$ $L = 270s$

Therefore, \frac{L}{s} = 270^irc\text{C}

5. Final Answer

The required ratio is: $\boxed{270}$