Water (2350 g) is heated until it just begins to boil. If the water absorbs 5.83*105 J of heat in the process, what was the initial temperature of the water? Express your answer with the appropriate units.

40.7062 °C

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C * ΔT

Where,

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g °C

m is the mass of water

ΔT = (100 - x) °C

Given,

Mass = 2350 g

Q = 5.83 * 10⁵ J

Applying the values as:

Q = m C * ΔT

5.83 * 10⁵ = 2350 * 4.184 * (100 - x)

x, Initial temperature = 40.7062 °C