Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.
Show all work -- you may receive partial credit!
Liquid nitrogen: boils at T(boil) = -196 Celsius
latent heat of vaporization = 2.00 x 10^5 J/kg
latent heat of fusion = 2.57 x 10^4 J/kg
Aluminum: thermal conductivity = 240 J/s*m*C
One must keep an electronic camera must be very cold to avoid noisy images. A camera is connected to a reservoir of liquid nitrogen to chill it. The liquid nitrogen is at its boiling temperature. An aluminum wire of length L = 3 meters and diameter d = 6 millimeters runs from the reservoir to the camera.
Question 1: Every hour, a mass of m = 43 g of nitrogen evaporates. What is the rate of evaporation, in kg/sec?
Answer: m = 43 g = 0.043 kg.
0.043 kg kg
rate = ------------ = 1.2 x 10^(-5) ---
3600 sec sec
Question 2: How much energy is required to flow into the liquid nitrogen per second to maintain this evaporation?
Answer:
Energy = (Latent heat of vaporization)*(mass)
Energy mass
------ = (Latent heat of vaporization)* ----
sec sec
= (2.00 x 10^5 J/kg) * 1.2 x 10^(-5) kg/s
= 2.4 J/s
Question 3: What is the temperature of the camera, at the other end of the wire?
Answer: The rate of heat transfer from the vat of nitrogen to the camera
is
Q A * (delta T)
--- = k * -------------
t L
The cross-section area of the wire is
A = pi*r^2 = pi*(0.003 m)^2 = 2.8 x 10^(-5) m^2
Solve for (delta T)
Q L
(delta T) = --- * -----
t k*A
3 m
= (2.4 J/s) * ----------------------------------
(240 J/s*m*C)*(2.8 x 10^(-5) m^2)
= 106 Celsius
Now we can determine the temperature of the camera
T(camera) = T(nitrogen) + (delta T)
= -196 Celsius + 106 Celsius = -90 Celsius
This page maintained by Michael Richmond. Last modified Jan 13, 1997.
Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.