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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.

Respuesta :

superman1987

Answer:

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.

Eduard22sly

The amount of heat added to the water to raise its temperature from 17.5 °C to 29.7 °C is 4.71×10¹⁰ J

How to determine the volume of water

  • Dimension = 20.0 m x 12.5 m
  • Height = 3.73 m
  • Volume =?

Volume = dimension × height

Volume = 20 × 12.5 × 3.73

Volume = 932.5 m³

Multiply by 10⁶ to express in mL

Volume = 932.5 × 10⁶

Volume = 9.235×10⁸ mL

How to determine the mass of water

  • Volume = 9.235×10⁸ mL
  • Density = 1 g/mL
  • Mass of water =?

Mass = Density × Volume

Mass of water = 1 × 9.235×10⁸

Mass of water = 9.235×10⁸ g

How to determine the heat  

  • Mass of water (M) = 9.235×10⁸ g
  • Initial temperature (T₁) = 17.5 °C  
  • Final temperature (T₂) = 29.7 °C
  • Change in temperature (ΔT) = 29.7  – 17.5  = 12.2 °C
  • Specific heat capacity (C) = 4.184 J/gºC
  • Heat (Q) =?

Q = MCΔT

Q = 9.235×10⁸ × 4.184 × 12.2

Q = 4.71×10¹⁰ J

Learn more about heat transfer:

https://brainly.com/question/10286596

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