A 10.0 cm by 4.00 cm pool is filled with water to a depth of 2.50 cm. How many joules are needed to heat the water from 18.5oC to 32.0oC? Cp of water = 4.184 J/goC

5648.4 Joules

Explanation:

This problem can be solved by the calorimetry formula:

Q = C . m . ΔT

where ΔT indicates the change of temperature

Final T°  - Initial T°

m = mass

C = specific heat of water and Q = Heat

We do not have m, but we have the volume of water. We can obtain mass from density.

Volume of water = 10 cm . 4cm . 2.50 cm = 100 cm³

1 cm³ = 1 mL

Density of water = 1 g/mL

Therefore our 100 mL of water is the volume for 100 g. We replace data:

Q = 4.184 J/g°C . 100 g . (32°C - 18.5°C)

Q = 5648.4 Joules