Answer:
a. 1.00 x 10⁴ J = 10.0 kJ
b. 1.42 x 10⁴ J = 14.2 kJ
Explanation:
Given the change in temperature during the reaction and assuming the volume of water and density remains constant, the change in enthalpy for the reaction will be given by
ΔHxn = Q = mCΔT where,
m= mass of water
C= specific heat of water, and
ΔT= change in temperature
a. mH₂O = 400.0 mL x 1.00g/mL = 400.00 g
Q = ΔHrxn = 400.00g x 4.184 J/gºC x 6.00 ºC = 1.00 x 10⁴ J = 10.0 kJ
b. mH₂O = 200.0 mL x 1.00 g/mL = 200.0 g
Q = ΔHrxn = 200.00 g x 4.184 J/gºC x 17.0ºC = 1.42 x 10⁴ J = 14.2 kJ