Question

10g of water at 10 degree celsius is mixed with 50 g of copper at 10 degree celsius ,find the final temperature of mixture when specific heat capacity of water is 4.2J/g degree celsius and of copper is 0.4J/g celsius

Answer 1

Answer:

ΔQ=mcΔT

for water : ΔQ₁= 10*4.2*(10-T)               ∵ T - equilibrium temperature

for copper : ΔQ₂= 50*0.4*(T-10)

heat given by water is absorbed by copper

∴ΔQ₁=ΔQ₂

   21-2.1T=T-10

        31 = 3.1T

          T=10

Explanation: