Question

The temperature of a system drops by 45°F during a cooling process. Express this drop in temperature in K, R, and °C.

Answer 1

Answer: The temperature drop in K is 280.37 K, in degree Celsius is 7.22°C and in Rankine is 504.67 R

Explanation:

We are given:

Temperature in degree Fahrenheit = 45°F

• To convert this temperature into degree Celsius, we use the conversion factor:

$^oC=(45^oF-32)\times \frac{5}{9}$

Putting values in above equation, we get:

$^oC=(45^oF-32)\times \frac{5}{9}\\\\T(^oC)=7.22^oC$

Temperature in degree Celsius = 7.22°C

• To convert this temperature into Kelvins, we use the conversion factor:

$(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}$

Putting values in above equation, we get:

$T(K)=(45^oF-32)\times \frac{5}{9}+273.15\\\\T(K)=280.37K$

Temperature in Kelvins = 280.37 K

• To convert this temperature into Rankine, we use the conversion factor:

$T(R)=T(^oF)+459.67$

Putting values in above equation, we get:

$T(R)=45^oF+459.67\\\\T(^oR)=504.67R$

Temperature in Rankine = 504.67 R

Hence, the temperature drop in K is 280.37 K, in degree Celsius is 7.22°C and in Rankine is 504.67 R