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AH
Carnegie Mellon University

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Problem 3

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

(a) $\Delta T=49.0 \mathrm{F}^{\circ} . \Delta T=\left(49.0 \mathrm{F}^{\circ}\right)\left(\frac{1 \mathrm{C}^{\circ}}{\frac{9}{5} \mathrm{F}^{\circ}}\right)=27.2 \mathrm{C}^{\circ}$
(b) $\Delta T=-100 \mathrm{F}^{\circ} . \Delta T=\left(-100.0 \mathrm{F}^{\circ}\right)\left(\frac{1 \mathrm{C}^{\circ}}{\frac{9}{5} \mathrm{F}^{\circ}}\right)=-55.6 \mathrm{C}^{\circ}$

## Discussion

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## Video Transcript

so problem three were simply converting a Fahrenheit to Celsius. We can say that temperature in Celsius will be equal to 5/9 times the temperature and Fahrenheit plus 32. So essentially the change in temperature in Celsius would be equal to 5/9 times the temperature in Fahrenheit final minus the temperature and Fahrenheit initial. So this would mean for part A. This would mean that the change in temperature Celsius would be equal to 5/9 times, 45 minus negative for and this is going to equal 27.2 degrees Celsius. For part B. The change in temperature and Celsius would be equal to 5/9 times negative 56 minus 44 and this is giving us a negative 55.6 degrees Celsius. That is the end of the solution. Thank you for watching.