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

someone explain to me plz

Question

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

someone explain to me plz

anonymous · Answer

Okay, so this is thermochemistry and thermodynamics stuff. 
q=mcT
For c you use the specific heat of water (the 4.18)
For m you use the mass of water I believe, either that or the total..
And q is heat
To find q you use deltaH=q/moles
And when you find q just plug that in with the 4.18 and mass of water and it should give you the temperature