How much heat (in kJ k J ) is evolved in converting 2.00 mol m o l of steam at 135 ∘C ∘ C to ice at -42 ∘C ∘ C ? The heat capaci
ty of steam is 2.01 J/(g⋅∘C) J / ( g ⋅ ∘ C ) , and that one of ice is 2.09 J/(g⋅∘C) J / ( g ⋅ ∘ C ) .
1 answer:
The total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
<h3>
Heat evolved in converting the steam to ice</h3>
The total heat evolved is calculated as follows;
Q(tot) = Q1(steam to boiling point) + Q2(boiling point to ice) +Q3(freezing to -42 ⁰C)
where;
Q = = mcΔθ
where;
- m is mass, (mass of water = 18 g/mol)
- c is specific heat capacity,
- Δθ is change in temperature
Q(tot) = 2(18)(2.01)(100 - 135) + 2(18)(2.01)(0 - 100) + 2(18)(2.09)(-42 - 0)
Q(tot) = -12,928.68 J
Thus, the total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
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