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

Answer:

water is in the vapor state,

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