What mass of water at 25 C must be allowed to come to thermal equilibrium with a

Question

What mass of water at 25 C must be allowed to come to thermal equilibrium with a 3 kg gold bar at 100C in order to lower the temp of the bar to 50 C.
found other versions of this on chegg and yahoo, which helped. But I'm still not getting the right answer. Q=mcdelta T. Since heat gained by water = heat lost by gold set q for gold to q for water.

(3kg)(126J/kg/C)(50) = m (4186 J/Kg/C)(75)
18900 = m (31390)
m = 0.6 kg , except its 190 grams. so obviously I'm still doing something wrong. Thanks for the help.

Explanation / Answer

heat gained = heat lost m1s1(T1-T)=m2s2(T-T2) 3*126*(100-50)=m*4186*(50-25) m=0.1806 kg

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