Question

If the heat necessary to warm 565.0 g of water from a temperature of T1 = 22.0 °C to T2 = 80.0 °C were somehow converted to translational kinetic energy of this amount of water, what would be the speed of this water?

Answer 1

696.83 m/s

Step-by-step explanation:

m = mass of water = 565 g = 0.565 kg

c = specific heat of water = 4186 J/(kg⁰C)

ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C

v = speed gained by water

Using conservation of energy

Kinetic energy gained by water = heat required to warm water

(0.5) m v² = m c ΔT

(0.5) v² = c ΔT

(0.5) v² = (4186) (58)

v = 696.83 m/s