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(a) On January 22, 1943, the temperature in Spearfish,

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 3E Chapter 17

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 3E

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from - 4.0o F to 45.0o F in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was 44.0o F on January 23, 1916. The next day the temperature plummeted to - 56o F. What was the temperature change in Celsius degrees?

Step-by-Step Solution:

Solution 3E Step 1 of 5: Let the two fahrenheit scale temperatures be T and T and corresponding celsius F1 F2 equivalent be T C1and T C2 respectively. Equation for converting temperature from fahrenheit to celsius is, T c (T9 F2 )…………….1 Where T and T are the temperatures in celsius and fahrenheit scale respectively. c F Equation to calculate the change in celsius scale, T C (final temperature) (initial temperature) T = T T ……………...2 C C2 C1 Step 2 of 5: (a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from - 4.0o F to 45.0o F in just 2 minutes. What was the temperature change in Celsius degrees Given data, Fahrenheit scale temperatures , T = 4 F and T = 45 F F1 F2 Corresponding celsius temperature, 0 For T F1 = 4 F Using in equation 1, T C1 = ( 4 32 )0 9 T = 20 c C1 0 For T F2 = 45 F Using in equation 1, T C2 = 945 32 )0 T = 7.22 c0 C2 Step 3 of 5: To calculate change in celsius scale, Substituting T C1 = 20 c and T C2 = 7.22 c in equation 2 , T = 7.22 c ( 20 c) 0 C 0 T C 27 .22 c Therefore, the change in celsius scale for the given case is 27 .22 c.

Step 4 of 5

Chapter 17, Problem 3E is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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