Question

Practice With Calorimetry And Heat 1. 75.0 g of cast iron was heated to 100°C and then plunged into 100 g of water. 23.0°C. Calculate the final temperature. Cast iron has specific heat of 0.46 19 2. A 25.0 g sample of an unknown metal at 99.5°C is placed into a calorimeter holding 50.0 g of water at 22.3°C. The final temperature was 26.2°C; what was the specific heat of the metal?

Answer 1

1. Heat lost by Cast iron = Heat gained by water

         m1*s1*DT1 = m2*s2*DT2

m1 = mass of cast iron = 75 g, s1 = specific heat of cast iron = 0.46 j/g.c, DT = (100-t)

m2 = mass of water = 100 g, s2 = specific heat of water = 4.184 j/g.c, DT2 = (23-t)

75*0.46*(100-t) = 100*4.184*(23-t)
The final temperature is 28.86 ⁰C

2.

heat lost by metal = heat gained by water

         m1*s1*DT1 = m2*s2*DT2

m1 = mass of metal = 25 g

s1 = specific heat of metal = ?

DT = 99.5-26.2 = 73.3 c

m2 = mass of water = 50 g

s2 = specific heat of water = 4.184 j/g.c

DT2 = 26.2-22.3 = 3.9 c

25*S1*73.3 = 50*4.184*3.9
S1 = specific heat of metal = 0.445 j/g.c