PHYS-212Spring 20162interior surface temperature is 19.0 oC, and the exterior su

Question

PHYS-212Spring 20162interior surface temperature is 19.0 oC, and the exterior surface temperature is -10.0 oC. (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall? [8points]6.The emissivity of tungsten is 0.350. A tungsten sphere with radius 1.50 cm is suspended within a large evacuated enclosure whose walls are at 290.0 K. What power input is required to maintain the sphere at a temperature of 3000.0 Kif the heat conduction along the supports is neglected? [5points]7.Helium gas with a volume of 2.60 L, under a pressure of 0.180 atm and at a temperature of 41.0 oC, is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 g/mol.

Explanation / Answer

2)

in the question the thickness of wood is not given assuming it as a units

a = thickness of the wood layer and styrofoam
kw = 0.08 W/mK thermal conductivity of wood
To = - 10°C = 263 K outside temperature

b thickness of styrofoam
ks = 0.01 W/mK thermal conductivity of styrofoam
Ti = 19°C = 292 K inside temperature

Tx = temperature at contact point wood - styrofoam

The rate of heat transfer must be the same through the wood and styrofoam layers :

ks ( Ti - Tx ) / b = kw ( Tx - To ) / a

Tx ( kw/a + ks/b) = ks Ti/b + kw To/a

Tx = ( 0.01*292/b + 0.08*263/a) / ( 0.08/a + 0.01/b) = 267 K

solve the above eqwuation for Tx

The rate of heat flow per square meter is :

= T / R ; where R is total thermal resistance of the wall R = Rw + Rs

Rw thermal resistance of wood layer Rw = a / A*kw = a / 0.08
Rs thermal resistance of styro layer Rs = b / A*ks = b / 0.01
A = 1 square meter

R = Rw+ Rs
T = 19 - (-10) = 29 K

solve for
= 29 / R

6)

The sphere both emits radiation and absorbs it from the surrounding walls .. so we use Stefan's law in the form ..

Net power emitted /m² ... P/m² = (Tsp^4 - Tw^4) .. .. Tsp= sphere temp, Tw= wall temp

P/m² = 0.350 x 5.67x10^-8 x (3000^4 - 290^4) = 16.07x 10^5 W/m²

Net power emitted = 16.07x 10^5W/m² x Area = net input power required.
Area = 4r² = 4 x x (0.0150m)² = 2.826x10^-3 m²

P =16.07x 10^5W/m² x 2.826x10^-3m² .. .. .. P = 4.54x 10^3 W .. (4.54kW)

7.  a) Using the gas equation
P1V1/T1 = P2V2/T2
Here let P1 = 0.18 atm so P2 = 2P1 atm; V1 = 2.60 L so V2 = 2V1 L; T1 = 273 + 41 = 314 K , T2 = ?
Thus T2 = P2xV2 x T1/P1 x V1 = 2P1 x 2V1 x 314/P1 x V1 = 1256 K OR 983.0 ^0C

b) To find the mass of He use the following equation
PV = nRT where n is the No. of moles of He
i.e. n = PV/RT
P = 0.18 atm, V = 2.60 L, T = 314 K, R = 0.0821 L atm. K^-1 mol^-1
So n = 0.18 x 2.60/0.0821 x 314 = 0.018
Thus the moles of He = 0.1018
And the mass of He = 0.018 x 4 = 0.0726g

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