Question

reasoning the formula $k = \frac{5}{9}(f - 32)+273.15$ converts temperatures from degrees fahrenheit $f$ to kelvin $k$.
a. solve the formula for $f$.
$f = $

b. the freezing point of helium is 0.95 kelvin. what is this temperature in degrees fahrenheit?
$0.95\\ k = square^{circ}f$

c. the temperature of dry ice is -78.5 $^{circ}c$. which is colder, dry ice or liquid nitrogen?

Explanation:

Step1: Isolate the term with \(F\)

\[

$$\begin{align*} K&=\frac{5}{9}(F - 32)+273.15\\ K- 273.15&=\frac{5}{9}(F - 32) \end{align*}$$

\]

Step2: Solve for \(F - 32\)

Multiply both sides by \(\frac{9}{5}\): \(\frac{9}{5}(K - 273.15)=F - 32\)

Step3: Solve for \(F\)

Add 32 to both sides: \(F=\frac{9}{5}(K - 273.15)+32\)

Step4: Find \(F\) when \(K = 0.95\)

\[

$$\begin{align*} F&=\frac{9}{5}(0.95 - 273.15)+32\\ &=\frac{9}{5}\times(- 272.2)+32\\ &=9\times(-54.44)+32\\ &=-489.96 + 32\\ &=-457.96 \end{align*}$$

\]

Answer:

a. \(F=\frac{9}{5}(K - 273.15)+32\)
b. \(-457.96\)
c. First, convert \(-78.5^{\circ}C\) to Kelvin: \(K=-78.5 + 273.15=194.65\)K. Since \(0.95\)K (freezing - point of helium) is less than \(194.65\)K (dry - ice temperature in Kelvin), the freezing - point of helium is colder. But the problem doesn't give the temperature of liquid nitrogen, so we can't fully answer this part with the given information. If we assume we just compare based on the dry - ice value we calculated, the answer related to dry ice: The freezing - point of helium (corresponding to \(0.95\)K or \(-457.96^{\circ}F\)) is colder than dry ice (\(-78.5^{\circ}C\) or \(194.65\)K).