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To find out the amount of heat required to cool the remaining water from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\), we can use the specific heat formula: Heat required (Q) = (mass of water to be cooled) \(\times\) (specific heat of water) \(\times\) (change in temperature) Let's denote the mass of water to be cooled as 'm'. Then, the change in temperature is \(35^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C}\). The specific heat of water is given as \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). Now, the formula becomes: Q = m \(\times 4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} \times 15^{\circ} \mathrm{C}\) #Step 3: Equate the heat lost due to evaporation to the heat required to cool the remaining water#

Now, we can solve the equation in Step 3 for the mass (m) of water that can be cooled: \[m = \frac{144000 \mathrm{~J}}{4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} \times 15^{\circ} \mathrm{C}}\] \[m = \frac{144000 \mathrm{~J}}{62.7 \mathrm{~J}/\mathrm{g}}\] \[m = 2296 \mathrm{~g}\] Hence, \(2296 \mathrm{~g}\) of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.