10.1 g CaO is dropped into a styrofoam coffee cup containing 157 g H2O at 18.0°C. If the following reaction occurs, then 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? [specific heat of water = 4.18 J/g·°C]

CaO(s) + H2O(l) → Ca(OH)2(s) triangleH^0(subcript rxn) = −64.8 kJ/mol

a. 18.02°C b. 35.8°C c. 311°C d. 42.2°C e. 117°C

We calculate the moles of CaO present:
Moles = mass / Mr
= 10.1 / 56
= 0.18 mol

1 mol of CaO releases 64.8 kJ of energy
0.18 mol will release:
64.8 x 0.18 = 11.664 kJ

To calculate the change in temperature of water:
Q = mCpΔT
Cp = 4.18 J/g or 4.18 kJ/kg

11.664 = 0.157 x 4.18 x (T₂ - 18)
T₂ = 35.8 °C

The answer is B.

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