What is the final temperature when 50 mL of water at 80°C are added to 25 mL of water at 25°C?

1 Answer

Answer:

#61.7^oC#

Explanation:

#sum# energy changes = 0

#sumQ = Q_(80^oC) + Q_(25^oC)# = #[mcDeltaT_(80^oC) + mcDeltaT_(25^oC)#] = 0

Density HOH = 1.00 g/ml;
Speific Heat(c) HOH = #1.00 "cal"/g^oC#

#[(50g)(1("cal"/g^oC))(T_("final") - 80^oC)]# +

#[(25g)(1("cal"/g^oC))(T_("final") - 25^oC)]# = #0#

=> #[(50)(1)(T_("final") - 80)] + [(25)(1)(T_("final") - 25)] = 0#

=> #(50T_("final") - 4000) + (25T_("final") - 625) = 0#

=> #75T_("final") = (4000 + 625) = 4625#

=> #T = (4625/75)^oC = 61.7^oC#