Question

How much heat is evolved in converting 1.00 mol of steam at 145.0 ∘C to ice at -55.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

Answer:

Heat evolved = 57.9 kJ i.e. -57.9 kJ

Explanation:

The transition from steam at 145 C to ice at -55 C involves the following steps:

Step 1: Heat involved in the change from steam at 145 C to steam at 100 C

$q1 = mc(T2-T1)\\\\m = mass\ of\ steam\\\\c = heat\ capacity\ of\ steam\\\\T2, T1 = final\ and\ initial\ temperatures$

Moles of steam (H2O)= 1 mole

Molar mass of H2O = 18 g/mol

$Mass\ of\ steam,\ m = moles*molar\ mass= 1 mole*18 g/mol= 18\ g\\\\c = 2.01 J/gC\\\\T2-T1 = 100-145 = -45 C\\\\q1 = 18 g*2.01J/gC *(-45C) = -1628.1J= -1.628\ kJ$

Step 2: Phase change from steam at 100 C to water at 100 C

$q2 = n\Delta H(cond)$

where ΔH(cond) = heat of condensation = -40.7 kJ/mol

n = number of moles$q2 = 1\ mole*(-40.7\ kJ/mol)= -40.7\ kJ$

Step 3: Heat involved in the change from water at 100 C to water at 0 C

$q3 = mc(T2-T1)\\\\m = mass\ of\ water\\\\c = heat\ capacity\ of\ water\\\\T2, T1 = final\ and\ initial\ temperatures$

$q3 = 18 g*4.18J/gC *(-100C) = -7524J= -7.524\ kJ$

Step 4: Phase change from water at 0 C to ice at 0 C

$q4 = n\Delta H(cond)$

where ΔH(fusion) = -6.01 kJ/mol

n = number of moles$q4 = 1\ mole*(-6.01\ kJ/mol)= -6.01\ kJ$

Step 5: Heat involved in the change from ice at 0 C to ice at -55 C

$q5 = mc(T2-T1)\\\\m = mass\ of\ water\\\\c = heat\ capacity\ of\ ice\\\\T2, T1 = final\ and\ initial\ temperatures$

$q5 = 18 g*2.09J/gC *(-55C) = -2069.1\ J= -2.069\ kJ$

Step 6: Total heat evolved

$q = q1+q2+q3+q4+q5$

q = -1.628 -40.7 -7.524-6.01-2.069 = -57.9 kJ