10.1 g CaO is dropped into a styrofoam coffee cup containing 157 g H₂O at 18.0^$\circ$C.If the following reaction occurs, what temperature will the water reach, assuming that the cup is a perfect insulator and that the cup absorbs only a negligible amount of heat (The specific heat of water = 4.18 J/g·^$\circ$C)
CaO(s) + H₂O(l) $\rarr$ Ca(OH) ₂(s) $\Delta$ H^$\circ$_rxn = -64.8 kJ/mol

Question

10.1 g CaO is dropped into a styrofoam coffee cup containing 157 g H₂O at 18.0^$\circ$C.If the following reaction occurs, what temperature will the water reach, assuming that the cup is a perfect insulator and that the cup absorbs only a negligible amount of heat (The specific heat of water = 4.18 J/g·^$\circ$C)
CaO(s) + H₂O(l)

\rarr

Ca(OH) ₂(s)

\Delta

H^$\circ$_rxn = -64.8 kJ/mol

Quizplus · Accepted Answer

First, calculate the moles of CaO used: $ \frac{10.1 \, \text{g}}{56.08 \, \text{g/mol}} = 0.180 \, \text{moles} $. The heat released by the reaction is $ 0.180 \, \text{moles} \times -64.8 \, \text{kJ/mol} = -11.664 \, \text{kJ} $. Convert this to joules: $ -11.664 \, \text{kJ} \times 1000 = -11664 \, \text{J} $. The heat absorbed by the water can be calculated using $ q = mc\Delta T $, rearranging for $ \Delta T $ gives $ \Delta T = \frac{q}{mc} $. Substituting the values gives $ \Delta T = \frac{-11664 \, \text{J}}{157 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C}} = -17.8^\circ\text{C} $. Adding this temperature change to the initial temperature of $ 18^\circ\text{C} $ results in a final temperature of $ 35.8^\circ\text{C} $.

101 G CaO Is Dropped into a Styrofoam Coffee Cup ∘\circ∘

101 G CaO Is Dropped into a Styrofoam Coffee Cup $\circ$