Question

If you have 0.426 m^3 of water at 25.0 ∘C in an insulated container and add 0.132 m^3 of water at 95.0 ∘C, what is the final temperature f of the mixture? Use 1000 kg/m^3 as the density of water at any temperature.

Answer 1

Answer: 41.47°c

Explanation:

0.426 m^3 * 1000 kg/m^3 = 426 kg

0.132 m^3 * 1000 kg/m^3 = 132 kg

T(f) = [(0.132 kg * 95°c) + (426 kg * 25°c)]/ (426 kg + 132 kg)

T(f) = 41.47°c