Problem Solving

Question

1. A fluid system has a temperature of 83^oF. What is its temperature expressed in Kelvin?

2. A Celsius and Fahrenheit thermometer are used to measure the temperature of a certain substance. If the temperature readings from both thermometers are numerically equal,

What is the temperature of the substance?

3. Consider a person whose exposed surface area is 2 m², emissivity is 0.5, and surface temperature is 32°C. Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of 490 R

Answer 1

1) I would first convert to degrees C using C = 5/9(F - 32) so C = 5/9(83 - 32) = 5/9•51 = 85/3 °C. Then to convert from °C to K, add 273.15 so K = 85/3 + 273.15 = 301.48 K. But we are given only 2 sig figs so to write this as 2 sig figs requires use of scientific notation. So the temperature in Kelvin is 3.0 x 10² K.

2) Let "x" be the temperature in °C then "x" is also the temp in °F according to the problem.

So x = 5/9(x - 32)

9/5x = x - 32

9/5x - x = -32

9/5x - 5/5x = -32

4/5x = -32

x = (-32)5/4

x = -40

So the temp of the substance is -40 °C or -40 °F

3) The radiation heat transfer equation looks like:

Q = εσA(T₁⁴ - T₂⁴) where ε is the emissivity, σ is the Stefan Boltzmann constant = 5.669 x 10^-8 W/(m²K⁴). We want to use Kelvin in the equation so first convert 32°C to K by adding 273.15 so the temperature of the person's skin is 305.15 K. To convert temp in R to K, first convert R to F by subtracting 459.67 so 490 R is 30.33 °F. Then convert °F to °C using C = 5/9(F - 32) = 5/9(30.33 - 32) = -0.93 °C. Then convert to K by adding 273.15. So the walls of the room are 272.22 K. That means heat will be transferred from the person's skin to the walls of the room.

Using the equation:

Q = εσA(T₁⁴ - T₂⁴) and plugging in the values we get:

Q = (0.5)(5.669 x 10^-8)(2)(305.15⁴ - 272.22⁴)

Q = (5.669 x 10^-8)(3179144888)

Q = 180 Watts