Question

Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.75×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

37.18°C.

Step-by-step explanation:

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

where, Q is amount of heat absorbed by water (Q = 5.75 x 10⁵ J).

m is the mass of water (m = 2190 g).

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

ΔT is the temperature difference = (final T - initial T = 100.0°C - initial T).

∴ Q = m.c.ΔT = 5.75 x 10⁵ J.

5.75 x 10⁵ J = (2190 g)(4.18 J/g°C)(100.0°C - initial T)

∴ (100.0°C - initial T) = (5.75 x 10⁵ J)/(2190 g)(4.18 J/g°C) = 62.81°C.

∴ initial T = (100.0°C - 62.81°C) = 37.18°C.