Question

HELP

1) A 400g sample of alcohol (c = 2.43 J/g°C) at 16°C is mixed with 400g
of water (c = 4.19 J/g°C) at 85°C. What is the final temperature of the
mixture?

Answer 1

Step-by-step explanation:

Given data:

Mass of alcohol = 400 g

Specific heat capacity of alcohol = 2.43 J/g°C

Initial temperature of alcohol = 16°C

Mass of water = 400 g

Specific heat capacity of water = 4.19 J/g°C

Initial temperature of water = 85°C

Final temperature of mixture = ?

Solution:

Equation:

m₁c₁ (T₂-T₁ ) = m₂c₂(T₂-T₁)

by putting values,

400 × 4.19 × (T₂ - 85°C) = 400 × 2.43 × (T₂ - 16°C)

1676 × (T₂ - 85°C) = 972 × (T₂ - 16°C)