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

aaronq · Answer

the heat released by the reaction (system) raises the temperature of the water (surroundings), 
                                 $q_{system}=-q_{surroundings}$

you use $q=m*C_p*\Delta T$;       where $\Delta T= T_f-Ti$


                                $q_{water}=m*C_p*(\color{red}{T_f}-Ti)$

you're solving for the final temp, $T_f$.

BUT first, convert 10.1 g CaO to moles, then find the heat released, $q_{system}$.

          $q_{system}=n_{CaO}*\Delta H_{rxn}$;           $ \Delta H_{rxn}$=−64.8 kJ/mol