Question

A lightning flash releases about 1010J of electrical energy. Part A If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37∘C, what are the final state and temperature of the water? The specific heat of water is 4180 J/kg⋅ ∘C, heat of vaporization at the boiling temperature for water is 2.256×106J/kg, the specific heat of steam is 1970 J/kg⋅ ∘C.

Answer 1

water is in the vapor state,

Step-by-step explanation:

We must use calorimetry equations to find the final water temperatures. We assume that all energy is transformed into heat

E = Q₁ +
`Q_(L)`

Where Q1 is the heat required to bring water from the current temperature to the boiling point

Q₁ = m
`c_(e)` (
`T_(f)` -T₀)

Q₁ = 50 4180 (100 - 37)

Q₁ = 1.317 10⁷ J

Let's calculate the energy so that all the water changes state

`Q_(L)` = m L

`Q_(L)` = 50 2,256 106

`Q_(L)` = 1,128 10⁸ J

Let's look for the energy needed to convert all the water into steam is

Qt = Q₁ +
`Q_(L)`

Qt = 1.317 107 + 11.28 107

Qt = 12,597 10⁷ J

Let's calculate how much energy is left to heat the water vapor

ΔE = E - Qt

ΔE = 10¹⁰ - 12,597 10⁷

ΔE = 1000 107 - 12,597 107

ΔE = 987.4 10⁷ J

With this energy we heat the steam, clear the final temperature

Q = ΔE = m
`c_(e)` (
`T_(f)`-To)

(
`T_(f)`-T₀) = ΔE / m
`c_(e)`

`T_(f)` = T₀ + ΔE / m
`c_(e)`

`T_(f)` = 100 + 987.4 10⁷ / (50 1970)

`T_(f)` = 100 + 1,002 10⁵

`T_(f)` = 1,003 10⁵ ° C

This result indicates that the water is in the vapor state, in realizing at this temperature the water will be dissociated into its hydrogen and oxygen components