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28. At 10:30 am., the temperature of a heated metal was 1200 °F. The metal was then cryogenically cooled until at 10:42 am it reached the temperature of -80°F. a) What was the change in temperature in °F? °C = ( °F – 32 ) x 5/9 b) What was the change in temperature in °C per second? (use the conversion formulas to help) °F = ( °C x 9/5) + 32

28. At 10:30 am., the temperature of a heated metal was 1200 °F. The metal was then cryogenically cooled until at 10:42 am it reached the temperature of -80°F. a) What was the change in temperature in °F? °C = ( °F – 32 ) x 5/9 b) What was the change in temperature in °C per second? (use the conversion formulas to help) °F = ( °C x 9/5) + 32

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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28. At 10:30 am., the temperature of a heated metal was 1200 °F. The metal was then cryogenically cooled until at 10:42 am it reached the temperature of -80°F. a) What was the change in temperature in °F? °C = ( °F – 32 ) x 5/9 b) What was the change in temperature in °C per second? (use the conversion formulas to help) °F = ( °C x 9/5) + 32
Transcribed Image Text:28. At 10:30 am., the temperature of a heated metal was 1200 °F. The metal was then cryogenically cooled until at 10:42 am it reached the temperature of -80°F. a) What was the change in temperature in °F? °C = ( °F – 32 ) x 5/9 b) What was the change in temperature in °C per second? (use the conversion formulas to help) °F = ( °C x 9/5) + 32
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