Water cooling by evaporation For many years of drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from $\pu{35 ^\circ C}$ to $\pu{20 ^\circ C}$ by the evaporation of $\pu{60 g}$ of water?
Heat of vaporisation of water in this temperature range is $\pu{2.4 kJ g-1}$. Specific heat of water is $\pu{4.18 J g-1 K-1}$.

I'm pretty confuse by what the question is asking for. This is my attempt for the evaporation of $\pu{60 g}$ of water:
$$q = \pu{60 g} \cdot \pu{2.4 kJ g-1} = \pu{144 J}$$
I then subbed this energy unit ($\pu{144 J}$) to the heat change formula to find the mass of water. Why am I wrong?
It helps to write down the formulas, before you plug in the numbers.

Heat you need to extract to cool the water:

$$Q=m_1 c_p\Delta T$$

Heat needed to evaporate water:

$$Q=m_2L_v$$

So:

$$m_1 c_p\Delta T=m_2L_v$$

So:

$$m_1=\frac{m_2L_v}{c_p \Delta T}$$

Now plug in the numbers.