Question

1.    If water is cooled from 50^OC to -5^OC, what is the temperature change on the Celsius, Fahrenheit and Rankine scales?

Answer 1

Change in celsius, $\Delta T = 50-(-5) = 55^\circ C$

Initial temperature in Fahrenheit, $T_1 = \frac{(50× 9)}{5} + 32 = 122 F$

final temperature in Fahrenheit, $T_2 = \frac{(-5× 9)}{5} + 32 = 23 F$

Change in Fahrenhiet, $\Delta T = 122-23 = 99^\circ F$

Initial Temperature in Rankine, $T_1 = \frac{50 × 9}{5} + 491.67 = 581.67 R$

final Temperature in Rankine, $T_2 = \frac{-5 × 9}{5} + 491.67 = 482.67 R$

Change in temperature, $\Delta T = 581.67-482.67 = 99^\circ R$