Question

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 capacity 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 ) .

Answer 1

The total quantity of heat evolved in converting the steam to ice is determined as  -12,928.68 J.

Heat evolved in converting the steam to ice

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 is heat evolved

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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