Question

Water (2470 g ) is heated until it just begins to boil. If the water absorbs 5.47x105 J of heat in the process,

what was the initial temperature of the water?
Express your answer with the appropriate units.

Answer 1

47.01 °C

Step-by-step explanation:

Q = mcΔt

Mass of water = 2470g

c or specific heat capacity of water = 4.18 J °C g

Q = 5.47 x 10⁵ J, which is expanded 547,000 J

No idea what Δt is, but we do know the water began to boil, and we know that water's boiling point is 100°C, so the final temperature had to be 100°C.

547,000 J = (2470) x (4.18) x (Δt)

Rearrange equation for Δt

(547,000) / (2470 x 4.18) = Δt

Δt = 52.98°C

But we're not done. We're trying to find the INITIAL TEMPERATURE of the water, not the TEMPERATURE CHANGE. We already have the final temperature, which is 100°C, and now we have how much the initial temperature rose by to get to 100°C.

All that we have to do now is 100 - 52.98 = 47.02°C

The initial temperature of the water is 47.02°C