THERMAL PROPERTIES OF MATTER QUIZ-24
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Q231. A wire of length 1 m and radius 10(-3) m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium, its temperature is 900 K while that of surrounding is 300 K. The resistivity of the material of the wire at 300 K is Ï€2×10(-8) ohm m and its temperature coefficient of resistance is 7.8×10(-3)/°C (Stefan’s constant σ = 5.68 ×10(-8) W/m2 K4)
The resistivity of wire at 900 K is nearly
Solution
231
231
Q232. A solid aluminium sphere and a solid lead sphere of same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium=900 J/kg°C and that of lead = 130 J/kg°C. The density of lead =104 kg/m3 and that of aluminum =2.7×103kg/m3. Assume that the emissivity of both the spheres is the same
The ratio of rate of fall of temperature of the aluminium sphere to the rate of heat loss from the lead sphere is
Solution
232(a) Heat emitted by the surface of sphere per unit time P_r'=σT4 (πR2) Since the radius of both spheres is equal, the rate of heat loss by aluminium sphere = rate of heat loss by lead sphere
232(a) Heat emitted by the surface of sphere per unit time P_r'=σT4 (πR2) Since the radius of both spheres is equal, the rate of heat loss by aluminium sphere = rate of heat loss by lead sphere
Q233. Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at 100°C and 0°C, respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere
When steady state condition is reached everywhere which of the following statement is true?
Solution
233(d) Under steady state conditions, the temperatures at all section in the system remain constant and maintain a constant temperature gradient for a given material. The temperature gradient in copper, aluminium and brass will not be same however, the rate of heat conducted across all sections whether in copper or aluminium or brass will be the same
233(d) Under steady state conditions, the temperatures at all section in the system remain constant and maintain a constant temperature gradient for a given material. The temperature gradient in copper, aluminium and brass will not be same however, the rate of heat conducted across all sections whether in copper or aluminium or brass will be the same
Q234. A thin copper rod of uniform cross section A square metres and of length L metres has a spherical metal sphere of radius r metre at its one end symmetrically attached to the copper rod. The thermal conductivity of copper is ε. The free end of the copper rod is maintained at the temperature T kelvin by supplying thermal energy form a P watt source. Steady state conditions are allowed to be established while the rod is properly insulated against heat loss from its lateral surface. Surroundings are at 0°C. Stefan’s constant =σ W/m2 K4
After the steady state conditions are reached, the temperature of the spherical end rod, T_S is
Solution
234 (d) The distance from the end A of the copper rod (where power is supplied) and centre O of the metal sphere is (L+r). Hence, in the steady state conditions, P=(KA(T-T_s))/(L+r) Which gives T_S=T-(P(L+r))/KA (The heat is transmitted up to centre of the spherical end, and the sphere loses energy by radiation out of its spherical surface)
234 (d) The distance from the end A of the copper rod (where power is supplied) and centre O of the metal sphere is (L+r). Hence, in the steady state conditions, P=(KA(T-T_s))/(L+r) Which gives T_S=T-(P(L+r))/KA (The heat is transmitted up to centre of the spherical end, and the sphere loses energy by radiation out of its spherical surface)
Q235. An immersion heater, in an insulated vessel of negligible heat capacity brings 10 g of water to the boiling point from 16°C in 7 min. Then
Power of heater is nearly
Solution
235 (b) In 7 min, temperature of 100 g of water is raised by (1000-16°C)=84°C. The amount of the heat provided by heater Q_W=C_W m_W ∆T=(1 cal/g°C)(100g)(84°C) =8.4×103 cal=(8.4×103×4.186) J ≃3.5×104 J Power of heater=Q_w/t_1 =(8.4×103 cal)/((7×60) s)=20 cal/s =(20×4.18) J/s=83.6 W≈84 W
235 (b) In 7 min, temperature of 100 g of water is raised by (1000-16°C)=84°C. The amount of the heat provided by heater Q_W=C_W m_W ∆T=(1 cal/g°C)(100g)(84°C) =8.4×103 cal=(8.4×103×4.186) J ≃3.5×104 J Power of heater=Q_w/t_1 =(8.4×103 cal)/((7×60) s)=20 cal/s =(20×4.18) J/s=83.6 W≈84 W
Q236. A body of area 0.8×10(-2) m2 and mass 5×10(-4) kg directly faces the sum on a clear day. The body has an emissivity of 0.8 and specific heat of 0.8cal/kg K. The surrounding are at 27°C. (solar constant =1.4 kW/m2)
The rate of rise of the body’s temperature is nearly
Solution
236 (b) We consider the leaf to be a black body. The rate of energy radiated at any instant (dQ/dt)_e=σeAT_0^4 (i) If m is the mass of the body and C is its specific heat then the heat gained dQ/dt=mc dT/dt (ii) Since intensity of sun beam is uniform, power incident on the body is SA heat absorbed (dQ/dt)_ab=SA e Rate of increases of temperature =((net power absorbed))/(thermal capacity of body) =((SAα-σAeT4))/mc Given σ=5.67×10(-8) J/s m2 K4 A=0.8×10(-2) m2;α=e=0.8 S=1.4×103 W/m2 T=300 K m=5×10(-4) kg C=0.8 kcal/kg K=0.8×4.2 kJ/kg K dT/dt=([1.4×103×0.8×10(-2)×0.8-5.67×10(-8)×0.8×10(-2)×0.8×(300)4])/(5×10(-4)×(0.8×4.2×103)) =((8.96-2.9))/1.68=3.6°C/s=3.6 K/s
236 (b) We consider the leaf to be a black body. The rate of energy radiated at any instant (dQ/dt)_e=σeAT_0^4 (i) If m is the mass of the body and C is its specific heat then the heat gained dQ/dt=mc dT/dt (ii) Since intensity of sun beam is uniform, power incident on the body is SA heat absorbed (dQ/dt)_ab=SA e Rate of increases of temperature =((net power absorbed))/(thermal capacity of body) =((SAα-σAeT4))/mc Given σ=5.67×10(-8) J/s m2 K4 A=0.8×10(-2) m2;α=e=0.8 S=1.4×103 W/m2 T=300 K m=5×10(-4) kg C=0.8 kcal/kg K=0.8×4.2 kJ/kg K dT/dt=([1.4×103×0.8×10(-2)×0.8-5.67×10(-8)×0.8×10(-2)×0.8×(300)4])/(5×10(-4)×(0.8×4.2×103)) =((8.96-2.9))/1.68=3.6°C/s=3.6 K/s
Q237. A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at 20°C. The inside diameter of the copper collar at that temperature is 5.98 cm
To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the steel shaft remains at 20°C?(α_copper=17×10(-6)/K)
Solution
237 (c) l_1=l_0 (1+αt) (2Ï€×6)/2=l_0 (1+αt) (2Ï€×5.98)/2=l_0 (1+20α) 6/5.98=(1+αt)/(1+20α)=((1+17×10(-6) t))/(1+20×17×10(-6) ) ∴t=216.8°C≈217°C
237 (c) l_1=l_0 (1+αt) (2Ï€×6)/2=l_0 (1+αt) (2Ï€×5.98)/2=l_0 (1+20α) 6/5.98=(1+αt)/(1+20α)=((1+17×10(-6) t))/(1+20×17×10(-6) ) ∴t=216.8°C≈217°C
Q238. Two insulated metal bars each of length 5 cm and rectangular cross section with sides 2 cm and 3 cm are wedged between two walls, one held at 100° C and the other at 0° C. The bars are made of lead and silver.〖 K〗_pb=350 W/mK,K_Ag=425 W/mK
Thermal current through lead bar is
Solution
238(b) Resistance of lead bar R_Pb=l_Pb/(K_Pb A_Pb )=(5×10(-2))/(350×6×10(-4) ) =10/21=5/21 K/W Thermal current through lead bar I_Pb=∆T/R_Pb =(100×21)/5=420 W
238(b) Resistance of lead bar R_Pb=l_Pb/(K_Pb A_Pb )=(5×10(-2))/(350×6×10(-4) ) =10/21=5/21 K/W Thermal current through lead bar I_Pb=∆T/R_Pb =(100×21)/5=420 W
