Question

If you add a 95 C gold piece with a mass of 2.34g (C=.129 J/gC) to 50g of 22.2 C water in a calorimeter that will absorb no heat. What is the final temperature of the water?

Answer 1

Since calorimeter will not absorb heat, q_gold + q _water = 0

We know that q = m x C x

Given

m _gold = 2.34 g

m_water = 50 g

C_gold = 0.129 Jg^{-1 o}C^-1

C_water = 4.184 Jg^{-1 o}C^-1

T_initail (gold) = 95 ^oC

T_initail (water) = 22.2 ^oC

q_gold + q _water = 0

[(2.34 g) x (0.129 Jg^{-1 o}C^-1 ) x (T_f - 95 ) ^oC ] + [(50 g) x (4.184 Jg^{-1 o}C^-1 ) x (T_f - 22.2 ) ^oC ] = 0

0.30186 (T_f - 95 ) + 209.2 x (T_f - 22.2 ) ] = 0

0.30186T_f - 28.6767 +  209.2T_f - 4644.24 = 0

209.50186 T_f = 4672.9167

T_f = 4672.9167/209.50186 =22.3 ^oC