Verified answer

In cooling from 50 to 20 oC, there is no change of phase. So we simple use : Q = m x delta T x Cp

Q (heat) = 1000-g x 30 oC x 4.184-J/g*oC = 125,520-J = 125.520-kJ or 126 kJ (3 sig fig allowed).

Oops, sorry, I'm accustomed to calculating energy in Joules. 4.184-J = 1 cal and 4184-J = 1 kcal so 125,520 J /4184-J/kcal = 30.0 kcal (3 sig fig).

change in temperature = 30 degrees

1 calorie = change of one degree per gram of water

30 degrees X 1000 calorie/degree = 30,000 calories or 30 kcal heat loss