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HOMEWORK 5 - HOMEWORK 5 If two objects are in thermodynamic...

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HOMEWORK 5 If two objects are in thermodynamic equilibrium, can’t be at diff temp Thermometer substance- undergoes some change when heated or cooled Heat is: energy transferred due to temp change When a particular constant volume thermometer is in thermal contact with water at triple point (276K), pressure is 5.55*10^4Pa. For thermometer 1K corresponds to change in pressure of about: ΔT=1K, p/t=constant = (p + Δp)/(t + Δt); so Δp=p/t*Δt ; so Δp=5.55*10^4/276.16=203Pa Heat has same units as work. Heat capacity: amount of heat energy required to raise temp by 1C or 1K Specific heat: amount of heat energy per unit mass required to raise temp by 1C/1K Cube of aluminum, edge length 15.9cm, aluminum density 2.7 times that of water (1g/cm^3) and specific heat .217 times water (1cal). When internal energy increases by 94000 cal, temp increases by: (a=edge) Δt=ΔEint/mc; m=ρa^3 so m=(2.7)(15.9)^3=10850 Δt=94000/(10850g*.217) = 40deg Object A, with heat capacity C A and temp T A , is placed in thermal contact with object B, with heat capacity C B B and temp T B . B The combination is thermally isolated. Final temperature of both objects is: (if objects thermally isolated, amount of heat one loses=other’s gain) Ca(Tf-Ta)=Cb(Tb-Tf) Tf=(CaTa + CbTb)/(Ca + Cb) The heat capacity of object B is twice that of object A. Initially A is at 200 K and B is at 400 K. They are placed in thermal contact and the combination is isolated. The final temperature of both objects is: (Ca200 + 2Ca400)/3Ca= Ca(Ta+2Tb)/3Ca = (Ta + 2Tb)/3 = 333K The specific heat of lead is 0.030 cal/g C ° . A quantity of 500 g of lead shot at 100 ° C is mixed with 100 g of water at 70 ° C in an insulated container. The final temperature of the mixture is: Ca=500g(.030)=15cal/C Cb=100g(1.cal/g)=100cal/C Tf=blah blah= 74deg
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A slab of material has area A , thickness L , and thermal conductivity k . One of its surfaces (P) is maintained at temperature T 1 and the other surface (Q) is maintained at a lower temperature T 2 . The magnitude of the rate of heat flow from P to Q is: Rate of heat flow is: H=-kA(ΔT/ΔX); ΔX=L and ΔT=T1-T2(pos) (minus sign=flow from high to low) The rate of heat flow through a slab is H . If the slab thickness is halved, its cross- sectional area is doubled, and the temperature difference across it is doubled, then the rate of heat flow becomes: =8H When the temperature of a copper penny is increased by 100 C ° , its diameter increases by 0.17%. The area of one of its faces increases by: ΔA=2(pid^2/4)Δd/d so 34% The coefficient of linear expansion of iron is 1.1 × 10 –5 per C ° . The volume of an iron cube, 4.00 cm on each edge, will increase by what amount if it is heated from 15 ° C to 72 ° C?
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