Question

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Answer 1

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$.

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$.

Explanation:

(a)  Expression for change in temperature is as follows.

        $|\Delta T| = |x - y|K$

                         = 15.1 K

                    = $|(x - 273.15) - (y - 273.15)|^{o}C$

                    = $|x - y|^{o}C$

                    = $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$.

(b)  Change in temperature from Celsius to Fahrenheit is as follows.

           F = 1.8C + 32

          C = $\frac{F - 32}{1.8}$

Since,   K = C + 273

or,    $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

         $\Delta F = 1.8 \Delta K$

                      = 1.8 (15.1)

                      = $27.18^{o}F$

or,                  = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$.

Answer 2

Answer:

a.15.1 degree Celsius

b.27.18 degree Fahrenheit

Explanation:

We are given that

Temperature,$\Delta T$=15.1 K

a.We have to find the magnitude of its temperature change in degree Celsius.

Let initial temperature=x K

Final temperature=y K

$\Delta T=y-x$=15.1 K

$^{\circ} C=K-273$

Using the formula

$\mid\Delta T\mid=\mid{(x-273)-(y-273)}\mid$

$\mid\Delta T\mid=\mid y-x\mid=15.1^{\circ} C$

b.$^{\circ}F=1.8^{\circ}C+32$

$1.8^{\circ} C=^{\circ}F-32$

$^{\circ} C=\frac{^{\circ}F-32}{1.8}=\frac{^{\circ}F}{1.8}-17.8$

$\Delta ^{\circ}C=\frac{\Delta ^{\circ}F}{1.8}$

$K=^{\circ}C+273$

$\Delta K=\Delta ^{\circ}C=\frac{\Delta ^{\circ}F}{1.8}$

$\Delta ^{\circ}F=1.8\Delta K=1.8\times 15.1=27.18^{\circ} F$