Question

a piece of metal weighing 49.047 was heated to 100 C and then put into 100ml of water( initially at 23.7 C). they came to an equilibrium temperature of 27.8 c. assuming no heat lost to environment what is the specific heat of the metal

Answer 1

Explanation:

q(metal) = q(H2O)

-(m, metal x C, metal x DT, metal) = m, H2O x C, H2O x DT, H2O

-(49.047 g x C, metal x (27.8°C - 100°C) = 100 g x 4.184 x (27.8 – 22.3)

3,541.2 x C, metal = 2,301.2

C, metal = 0.65