Question

30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 343 K. What is the final temperature of the mixture

Answer 1

assuming density of water = 1 g/mL

mass water = 30.0 mL x 1 g/mL = 30.0 g
mass water = 50.0 mL x 1 g/mL = 50.0 g

specific heat water = 4.18 J/g°C

heat lost + heat gained = 0

30.0 g x 4.18 ( T - 282 ) + 50.0 x 4.18 ( T - 320)=0

solving for T

T = 305.6 K

Answer 2

The final temperature of the mixture will be "305.6 K".

Heat and temperature

According to the question,

Mass of water, m₁ = 30.0 mL or,

                              = 30.0 mL × 1 g/mL

                              = 30.0 g

and,

                        m₂ = 50.0 mL × 1 g/mL

                              = 50.0 g  

Specific heat of water = 4.18 J/g°C

We know the relation,

→ Heat lost + Heat gained = 0

By substituting the values,

  30.0 × 4.18 (T - 282) + 50.0 × 4.18 (T - 320) = 0  

30.0 × 4.18T - 1178.76 + 50.0 × 4.18T - 1337.6 = 0

hence,

The final temperature will be:

                                        T = 305.6 K

Thus the answer above is appropriate.

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