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How much heat in kilojoules is evolved in converting 1.00 mol of steam at 145 °C to ice
Chapter 12, Problem 95P(choose chapter or problem)
How much heat in kilojoules is evolved in converting 1.00 mol of steam at \(145^{\circ} \mathrm{C}\) to ice at \(-50.0^{\circ} \mathrm{C}\)? The heat capacity of steam is \(1.84 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\) and that of ice is \(2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\).
Equation Transcription:
Text Transcription:
145 degrees C
-50.0 degrees C
1.84 J/g degrees C
2.09 J/g degrees C
Questions & Answers
QUESTION:
How much heat in kilojoules is evolved in converting 1.00 mol of steam at \(145^{\circ} \mathrm{C}\) to ice at \(-50.0^{\circ} \mathrm{C}\)? The heat capacity of steam is \(1.84 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\) and that of ice is \(2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\).
Equation Transcription:
Text Transcription:
145 degrees C
-50.0 degrees C
1.84 J/g degrees C
2.09 J/g degrees C
ANSWER:
Solution 95P
Step 1:
Converting 1.00 mol of steam at 145 °C to ice at -50.0 °C involves the loss of
57.71 kJ of energy.
There are five heats to consider:
q1 = heat lost on cooling steam from 145 °C to 100 °C.
q2 = heat lost on condensing steam to water at 100 °C.
q3 = heat lost on cooling water from 100 °C to 0°C.
q4 = heat lost on freezing water to ice at 0 °C.
q5 = heat lost on cooling ice from 0 °C to -50.0 °.
The total heat evolved is
q=q1+q2+q3+q4+q5