Question 15 During the winter you visit the snowy mountains for a skiing holiday
ID: 1486628 • Letter: Q
Question
Question 15
During the winter you visit the snowy mountains for a skiing holiday. When you wake up one morning it is a chilly -4.17 oC outside. You put on a thick layer of clothing. You estimate the thickness of your clothes to be 2.30 cm and your surface area to be 1.98 m2. If your body temperature is 37.0 oC and you are loosing heat at a rate of 430 W what is the thermal conductivity of your clothing?
k = (Answer) Wm-1K-1
Question 16
To quantify the effectiveness of clothing as a thermal insulator clothing manufacturers have come up with a unit for “clothing insulation” the clo. 1 clo = 0.155 Km2W-1. You can read more about this unit on Wikipedia here: https://en.wikipedia.org/wiki/Clothing_insulation . Insulation and conductivity are inversely related to each other (something which is an insulator conducts hear slowly). The thermal insulation is related to the thermal conductivity through:
What is the clothing insulation of these clothes?
Clothing insulation = (Answer) clo
Question 17
While skiing you fall over a few times and end up fairly wet. How would this affect your clothing insulation?
Select one:
-It would decrease the clothing insulation.
-The clothing insulation would remain the same.
-It would increase the clothing insulation.
Question 18
You can see on the data sheet that 1 mole of an ideal gas at 0 oC and atmospheric pressure takes up a volume of 2.241 x 10-2 m3. At the top of the ski slope the pressure is measured to be 0.793 atm, the temperature is still -4.17 oC. What is the volume of exactly 1 mole of ideal gas under these conditions?
(Answer) m3
Explanation / Answer
Q 15
Solution: Temperature outside TC = -4.17 oC = (273.15 + -4.17) K = 268.98 K
Temperature of the body TH = 37.0 oC = (273.15 + 37.0) K = 310.15 K
Surface area of cloths A = 1.98 m2
Thickness of cloths, L = 2.30 cm = 0.023 m
Rate of loss of heat through the cloths, P = 430 W
The formula for heat conducted through object is,
P = k*A*(TH –TC)/L
k = P*L/ A*(TH –TC)
k = (430W)*(0.023m) / [(1.98m2)*(310.15K – 268.98K)
k = 0.12132 Wm-1K-1
The thermal conductivity of the clothing is 0.12 Wm-1K-1
-----------------------------------------------------------------------------------------------------------------------------------------
Q 16
Solution:
Thermal insulation = thickness / thermal conductivity
Thermal insulation = L/k
Thermal insulation = (0.023 m)/( 0.12132 Wm-1K-1)
Thermal insulation = 0.18957 Km2W-1
But 1 clo = 0.155 Km2W-1
Thermal insulation = (0.18957 Km2W-1)*( 1 clo / 0.155 Km2W-1)
Thermal insulation = 1.2231 clo
Hence the clothing insulation = 1.223 clo
----------------------------------------------------------------------------------------------------------------------------------------
Q 17
Solution: We assume that that the clothing do not allow water to leak through. The thermal conductivity k is a constant for that clothing material as long as its thickness and surface area remain the same. If the surrounding due to contact with water) temperature changes, the rate of heat loss will change accordingly but thermal conductivity k will remain the same. Since k does not change, the clothing insulation will not change too.
Hence we see that the clothing insulation would remain the same.
---------------------------------------------------------------------------------------------------------------------------------------------
Q 18
Solution: From the standard data,
At standard temperature pressure condition, for 1 mole of an ideal gas
The temperature Ts = 0 oC = 273.15 K
The pressure Ps = 1 atm
The volume Vs = 2.241*10-2 m3
Now at the top of the ski slope,
The pressure is P = 0.793 atm
The temperature T = -4.17 oC = (273.15 + -4.17) K = 268.98 K
We need to find the volume under these conditions.
From the ideal gas equation P*V =n*R*T,
n*R = P*V/T
Also for at the standard conditions,
n*R = Ps*Vs/Ts
Since the n*R is constant, we have
Ps*Vs/Ts = P*V/T
Vs = (P*V/T)*( Ts / Ps)
Vs = [(1atm)*( 2.241*10-2 m3)/(273.15 K)] / [(268.98 K)/(0.793 atm)]
Vs = 2.783*10-2 m3
Hence the volume of 1 mole of an ideal gas under these conditions is 2.783*10-2 m3