General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 1 - Matter: Its Properties and Measurement - Exercises - Temperature Scales - Page 28: 45

Answer

(a) $35.14 \space ^{\circ} M$ (b) $-59.27 ^{\circ} M$

Work Step by Step

$$T(^{\circ}M) = a \space T(^{\circ}C) + b$$ $$0 =a(-38.9) + b$$ $$38.9 \space a = b$$ $$T(^{\circ}M) = a \space T(^{\circ}C) + b $$ $$100 = a\space (356.9) + b$$ $$100 = 356.9 \space a + (38.9 \space a)$$ $$100 = 395.8 \space a$$ $$a = \frac{100}{395.8} \approx 0.253$$ $$b = 38.9 \times (0.253) \approx 9.84 $$ $$T(^{\circ} M) = 0.253 \space T(^{\circ} C) + 9.84$$ (a) $$T(^{\circ} M) = 0.253 \times (100) + 9.84 = 35.14$$ (b) $$T(^{\circ} M) = 0.253 \times (-273.15) + 9.84 = -59.27$$
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