Ask Question
16 October, 04:56

A swimming pool, 20.0 m ✕ 12.5 m, is filled with water to a depth of 3.73 m. If the initial temperature of the water is 17.5°C, how much heat must be added to the water to raise its temperature to 29.7°C? Assume that the density of water is 1.000 g/mL.

+3
Answers (1)
  1. 16 October, 05:21
    0
    4.755 x 10¹⁰ J.

    Explanation:

    The amount of heat absorbed by water = Q = m. c.ΔT.

    where, m is the mass of water (m = d x V).

    c is the specific heat capacity of liquid water = 4.18 J/g°C.

    ΔT is the temperature difference = (final T - initial T = 29.7°C - 17.5°C = 12.2°C).

    To calculate the mass of water in the swimming pool, we need to calculate its volume:

    Volume of the swimming pool = 20.0 m x 12.5 m x 3.73 m = 932.5 m³ x (10⁶ mL/1.0 m³) = 9.235 x 10⁸ mL.

    ∴ The mass of water in the swimming pool = d x V = (1.000 g/mL) (9.235 x 10⁸ mL) = 9.235 x 10⁸ g.

    ∴ The amount of heat absorbed by water = Q = m. c.ΔT = (9.235 x 10⁸ g) (4.18 J/g°C) (12.2°C) = 4.755 x 10¹⁰ J.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A swimming pool, 20.0 m ✕ 12.5 m, is filled with water to a depth of 3.73 m. If the initial temperature of the water is 17.5°C, how much ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers