J.R. S. answered 10/23/20
Ph.D. University Professor with 10+ years Tutoring Experience
a) q = mC∆T
q = heat = ?
m = mass = 10.9 g
C = specific heat = 0.916 J/gº
∆T = change in temperature = 37.9 - 22.2 = 15.7º
q = (10.9 g)(0.916 J/gº)(15.7º) = -157 J (negative sign because temperature went up -> exothermic rx)
b) q = mC∆T
q = 907 J
m = mass = 21.8 g
C = 0.916 J/gº
∆T = change in temp. = ?
Solve for ∆T = q/mC = 907 J/(21.8 g)(0.916 J/gº) = 4.54º
Since heat was added, the temperature will rise by this amount and final temp will be 22.2 + 4.54 = 26.7º