Question

If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state 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

Vapors

Step-by-step explanation:

We take into account that all the energy from the lightning has been transformed into steam.

`\Delta U = Q - W\\Q = mC \Delta T\\Q = mL`

We calculate the amount of energy required by water to convert into steam.

`Q_(water) = 50 * * 4180 * (100-37)\\= 1.132 * 10^7 \ J`

`Q_(change\ to\ steam) = 50 * 2.256 * 106 \\= 1.128 * 10^8 \ J`

`Q_(total) = 1.132 * 10^7 + 1.128 * 10^8\\= 1.126 * 10^8 \ J`

From the lightning we received
`10^(10) \ J` of energy, out of which
`1.126 * 10^8` has been used to convert the water into steam.

Energy left =
`10^(10) - 1.126 * 10^8 = 9.88 * 10^9 \ J`

We use this energy to convert steam into vapors.

`Q = \Delta E`

`Q = \Delta E = mc (T_(f) - T{i})\\T_(f) = \frac {\Delta E}{mc} + T_(i)\\ \\T_(f) = 100 + (9.88 * 10^(10))/(50 * 1970)\\T_(f) = 100 + 10^8\\T_(f) = 10 ^( \ 8) \ {^ \circ } C`

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.