Question

Path: P Words:0 QUESTION 27 A 42.9 g sample of Fe (specific heat 0.449 J/gºC) at an initial temperature of 125.0°C is added to 185.0 g sample of water. The final temperature reached for both is 42.1°C. Assuming no heat is lost to the surroundings and given that the specific heat of water is 4.18J/g°C, what is the initial temperature of the water, in °C? Simply type the number, not the unit. QUESTION 28

Answer 1

-mmetalx   Cmetal x ∆T = mwater x Cwater x ∆T                             i)

Given that

mmetal = 42.9 g

∆Tmetal = (42.1–125.0) =    - 82.90C

Cmetal = 0.449 J/0C g

Mwater =185.0g

∆Tmetal = (42.1 – Tinitial)oC

Cwater = 4.18 J/0C g

Put in eqn i)

- (42.9g) x 0.449 J/0C g x (-82.90C ) = 185 g x 4.18 J/0C g x (42.1 – Tinitial)oC

1596.83 / 773.3 = 42.1 – Tinitial

2.0649 = 42.1 – Tinitial

Tinitial = 42.1 – 2.0649 = 40.035 oC