Question

Water (2150 gg ) is heated until it just begins to boil. If the water absorbs 5.91×105 JJ of heat in the process, what was the initial temperature of the water?

Answer 1

Answer:

Initial temperature of water was 34,3°C

Explanation:

The heat absorbed in the process (Q) can be written as:

Q = mcΔT

Where the heat is 5,91x10⁵J, m is mass of water (2150g), c is specific heat of water (4,184J/g°C). And ΔT is change in temperature (100°C - x). Replacing:

5,91x10⁵J = 2150g×4,184J/g°C×(100°C - x)

5,91x10⁵J = 899560J - 8995,6J/°Cx

-3,0856x10⁵J = - 8995,6J/°Cx

34,3°C = x

Thus, initial temperature of water was 34,3°C

I hope it helps!