Answer to Question #154803 in Physical Chemistry for Aliza Huntoon

Question

How to calculate how much room temperature water (23.3°C) will be needed to drop the temperature of warm water from 50°C to 42°C.

Answer 1

Δ₁ = 50 – 42 = 8⁰C – cooling of warm water

Q₁ = cm₁Δ₁ = 8cm₁

Δ₂ = 42 – 23.3 = 18.7⁰C – heating up of room temperature water

Q₂ = cm₂Δ₂ = 18.7cm₂

Q₁ = Q₂

8cm₁ = 18.7cm₂

8m₁ = 18.7m₂

m₁ = 18.7m₂ / 8 = 2.34m₂

So, the mass of colder water mast be 2.34 times greater that the mass of warmer water.