Name Group GPI Recitation #13, 2017.11.28 You are helping a friend who is a vete
ID: 1794647 • Letter: N
Question
Name Group GPI Recitation #13, 2017.11.28 You are helping a friend who is a veterinarian to do some minor surgery on a cow in a remote location. Lacking a portable autoclave, she has asked you to sterilize a scalpel and a hemostat by placing them in boiling water for 30 minutes. You boil them as ordered in 2.5 kg of water, and then quickly transfer the instruments to a well insulated tray containing 250 grams of sterilized water at room temperature (23 C). After a few minutes the instruments and water will come to the same temperature- but will they be safe to hand to your friend without being burned? pel and the 75 gram hemostat are made from steel which You know that both the 45 gram scal has a specific heat of 450 J/(kg C), and the speciic heat of water at 23 °C is 4180 3/kg C) Assurningthat the temperature of the scalpel and hemostat are 100C after 30 minutes of ode., boiling, what is the final temperature of the instruments? The thermal conductivity of stainless steel is about 20 W/(m C), and each instrument is very roughly 10 cm long and 1 cm in diameter. ls it reasonable to assume the steel instruments will be 100 C after 30 minutes of boiling?Explanation / Answer
given mas sof scalpel, ms = 45 gram
and mh = 75 gram
specific heat of steel, c = 450 J /kg C
specific heat of water, C = 4180 J / kg C
final temeprature of the instruments after time t = T
now,
from newtons law of cooling
dT/dt = -k(T - 23)
k = hA/C
h = heat transfer coefficient = thermal conductivity / length
l = 10 cm
tc = 20 W/mC
hence
h = 20/0.1 = 200
A = pi*d*l = 0.0031415 m^2
C = 450
hence
k = 0.001396
hence
dT/(T - 23) = -kdt
integrating from 100 C to T
ln((T - 23)/(100 - 23)) = -kt
T = 23 + 77e^(-kt)
so after 10 minutes = 600 s
T = 56.31632 C
after 20 minutes
T = 37.4198
hence after 20 minutes the instruments are safe to handle
also,
dT/(T - 100) = -0.001396t
integrating from T = 23C to T
ln((T - 100)/(23 - 100)) = -0.001396t
so in t = 30 minutes
T = 93.75984 C
hence its reasonable to assume that in 30 minutes the temperature of the instruments can reach 100 C
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