If 55.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 55.0 mL of water (density = 1.0 g/mL) initially at 28.4 ∘C in an insulated beaker, what is the final temperature of the mixture, assuming that no heat is lost? (CEtOH=2.42J/(g⋅∘C).) Express the temperature in degrees Celsius to three significant figures.

T = 45.509 °C

Explanation:

EtOH:

∴ m EtOH = (55.0 mL)*(0.789 g/mL) = 43.395 g

∴ CEtOH = 2.42J/g°C

∴ T1 = 8.0 °C

H2O:

∴ m H2O = (55.0mL)*(1.0 g/mL) = 55 g

∴ C H2O = 4.186 J/g°C

∴ T1 = 28.4 °C

final temperature of the mixture (T):

⇒ Q EtOH = Q H2O

⇒ (43.395 g)(2.42 J/g°C)(T - 8.0°C) = (55.0 g)(4.186 J/g°C)(T - 28.4°C)

⇒  105.016 T - 840.127 = 230.23 T - 6538.532

⇒ 6538.532 - 840.127 = 230.23 T - 105.016 T

⇒ 5698.405J = (125.214 J/°C)( T )

⇒ T = 45.509 °C

it's c (engineer's do their work with mechanical stuff not on any living being)