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

Answer:

Vapors

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.

$Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J$

$Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J$

$Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J$

From the lightning we received $10^{10} \ J$ of energy, out of which $1.126 \times 10^8$ has been used to convert the water into steam.

Energy left = $10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J$

We use this energy to convert steam into vapors.

$Q = \Delta E$

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

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