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Thermal Properties of Matter Test - 45

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Thermal Properties of Matter Test - 45
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  • Question 1
    1 / -0
    A cubical box of side 0.20 m0.20\ m is constructed from 0.012 m0.012\ m thick concrete panels of thermal conductivity K=0.8 W m1K1K = 0.8\ W\ m^{-1} K^{-1}. A 100 W100\ W heater is sealed inside the box and switched on. The temperature of air outside the box is 20C20^{\circ}C. After a long period of time, the air temperature inside the box will be close to
    Solution
    Let the air temperature inside the box be xoCx^oC
    So,
    x20=100×0.0126×0.2×0.2×0.8{H=TlkA T=HlkA }x20=6.25x=26.25x26x-20=\cfrac { 100\times 0.012 }{ 6\times 0.2\times 0.2\times 0.8 } \quad \quad \quad \left\{ H=\cfrac { \triangle T }{ \cfrac { l }{ kA }  } \Rightarrow \triangle T=\cfrac { Hl }{ kA }  \right\} \\ \Rightarrow x-20=6.25\\ \Rightarrow x=26.25℃\\ \Rightarrow x\approx 26℃
  • Question 2
    1 / -0
    Vijay performed an experiment to compare the heat conductivity of four different materials P, Q, R and S. Thumbtacks were fixed with wax on each material and then the materials were fixed to a trough containing hot water as shown in the diagram. Vijay tabulated his results as shown here.
    MaterialTime taken for thumbtacks to drop(minutes)
    P55
    Q66
    R22
    S44
    What can be inferred from his experiment?

    Solution
    Material R is better conductor of heat because it took only 22 minutes for the wax attached to it to melt as compared to material S which took 44 minutes.
  • Question 3
    1 / -0
    two bars of same length and same cross-sectional area but of different thermal conductivities  K1 { K }_{ 1 } and  K2 { K }_{ 2 } are joined end to end as shown in the figure. One end of the compound bar is at temperature  T1 { T }_{ 1 } and the opposite end at temperature  t2 { t }_{ 2 } (where  T1>T2 { T }_{ 1 }>{ T }_{ 2 } ). the temperature of the junction is 

    Solution
    Let L and A be length and area of cross-section of each bar respectively.
      \therefore  Heat current through the bar 1 is
    H1=K2A(T1T0) L  { H }_{ 1 }=\dfrac { { K }_{ 2 }A\left( { T }_{ 1 }-{ T }_{ 0 } \right)  }{ L } 

    Here T0 { T }_{ 0 } is junction temperature.
    Heat current through the bar 2 is 
    H1=K2A(T0T2) L  { H }_{ 1 }=\dfrac { { K }_{ 2 }A\left( { T }_{ 0 }-{ T }_{ 2 } \right)  }{ L } 
    at steady state, H1=H2 { H }_{ 1 }={ H }_{ 2 }

    K1A(T1T0) L=K2A(T0T2) L  \therefore \quad \dfrac { { K }_{ 1 }A\left( { { T }_{ 1 } }-{ T }_{ 0 } \right)  }{ L } =\dfrac { { K }_{ 2 }A\left( { T }_{ 0 }-{ T }_{ 2 } \right)  }{ L } 

    K1(T1T0)=K2(T0T2)  { K }_{ 1 }\left( { T }_{ 1 }-{ T }_{ 0 } \right) ={ K }_{ 2 }\left( { T }_{ 0 }-{ T }_{ 2 } \right) 
    K1T1K1T0=K2T0K2T2 { K }_{ 1 }{ T }_{ 1 }-{ K }_{ 1 }{ T }_{ 0 }={ K }_{ 2 }{ T }_{ 0 }-{ K }_{ 2 }{ T }_{ 2 }
    K1T0K2T0=K1T1K2T2 { K }_{ 1 }{ T }_{ 0 }-{ K }_{ 2 }{ T }_{ 0 }={ K }_{ 1 }{ T }_{ 1 }-{ K }_{ 2 }{ T }_{ 2 }
    T0(K1+K2)=K1T1+K2T2 { T }_{ 0 }\left( { K }_{ 1 }+{ K }_{ 2 } \right) ={ K }_{ 1 }{ T }_{ 1 }+{ K }_{ 2 }{ T }_{ 2 }

    T0=K1T1+K2T2(K1+K2)   { T }_{ 0 }=\dfrac { { K }_{ 1 }{ T }_{ 1 }+{ K }_{ 2 }{ T }_{ 2 } }{ \left( { K }_{ 1 }+{ K }_{ 2 } \right)  } 

  • Question 4
    1 / -0
    One Kg of a diatomic gas is at a pressure of 8×1048\times 10^4N/ m2m^{2}. The density of the gas is 44kg/ m3m^{3}. The energy of the gas due to its thermal motion will be
    Solution
    For a diatomic molecular gas, internal energy per molecule is
    U=52NKBTU=\dfrac{5}{2}NK_BT
    PV=NKBTPV=NK_BT
    From equation is (i) and (ii)
    U=52U=\dfrac{5}{2}PV
    Here, P=8×104NP=8\times 10^4N m2,ρ=4kgm^{-2}, \rho =4kg m3m^{-3}, m=1m=1kg
    U=52×8×104×14=5×104\therefore U=\dfrac{5}{2}\times 8\times 10^4\times \dfrac{1}{4}=5\times 10^4J
  • Question 5
    1 / -0
    Which one of the following is not an assumption of kinetic theory of gases?
    Solution
    The correct answer is option (A).

    Explanation: The assumptions of the kinetic theory of gases are:
    1. Gases consist of a large number of molecules.
    2. Molecules of gases move rapidly and randomly with different speeds
    3. The volume of each molecule of gas is negligible compared to the volume occupied by the gas itself.
    4. Elastic collisions occur when one molecule collides with another or when they collide with the walls of the container.
    5. There is no force of attraction or repulsion between gas molecules.
    6. The average kinetic energy of gases depends on the Kelvin temperature.

    Hence, we see that all molecules do not have same speed according to the assumption of kinetic theory of gases.
    The correct answer is option (D).
  • Question 6
    1 / -0
    If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature
    Solution
    According to ideal gas equation
    P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2} or T2=P2V2P1V1T_2=\dfrac{P_2V_2}{P_1V_1}
    Here, P1=P,V1=V,T1=T,P2=P2,V2=V2,T2=?P_1=P, V_1=V, T_1=T, P_2=\dfrac{P}{2}, V_2=\dfrac{V}{2}, T_2=?
    T2=T(P2)(V2)PV\therefore T_2=\dfrac{T\left(\dfrac{P}{2}\right)\left(\dfrac{V}{2}\right)}{PV}, T2=T4T_2=\dfrac{T}{4}.
  • Question 7
    1 / -0
    Two blocks with heat capacities  C1 { C }_{ 1 } and  C2 { C }_{ 2 } are connected by a rod of length l, cross-sectional are A and heat conductivity K. Initial temperature difference between the two blocks is  T0 { T }_{ 0 } . Assuming the entire system to be isolated from surroundings, heat capacity of the rod to be negligible. The temperature difference between the blocks as a function of time is. 
    Solution

    Let T be the temperature difference between two blocks at time t.
    Heat transferred per second,
    dQdt=KATl  \dfrac { dQ }{ dt } =\dfrac { KAT }{ l }            \dots  (i)
    Also, dT=dT1+dT2 dT={ dT }_{ 1 }+{ dT }_{ 2 }       \dots  (ii)
    Heat lost by one block is equal to the heat gained by the 
    other, C1dT1=C2dT2 { C }_{ 1 }{ dT }_{ 1 }={ C }_{ 2 }{ dT }_{ 2 }                           \dots  (iii)
    From equations, (ii) and (iii), we get 
    dT=(C1+C2C2 )dT1 dT=\left( \dfrac { { C }_{ 1 }+{ C }_{ 2 } }{ { C }_{ 2 } }  \right) { dT }_{ 1 }                 \dots  (iv)
    If a block loses heat, dQ=C1dT1 dQ=-{ C }_{ 1 }{ dT }_{ 1 }
    From equation (i), dQdt=C1dT1dt=KATl  \frac { dQ }{ dt } =-{ C }_{ 1 }\frac { { dT }_{ 1 } }{ dt } =\dfrac { KAT }{ l }           (v)
    From equations (iv) and (v), we get
    C1C2C1+C2dTdt=KAlT \dfrac { -{ C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } \dfrac { dT }{ dt } =\dfrac { KA }{ l } T
    dTT=KA(C1+C2) C1C2dt \dfrac { dT }{ T } =\dfrac { -KA\left( { C }_{ 1 }+C_{ 2 } \right)  }{ { C }_{ 1 }{ C }_{ 2 } } dt
    or T0TdTT=KA(C1+C2) C1C20tdt   \int _{ { T }_{ 0 } }^{ T }{ \dfrac { dT }{ T } =\dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right)  }{ { C }_{ 1 }{ C }_{ 2 } } \int _{ 0 }^{ t }{ dt }  } 
    In TT0=KA(C1+C2) C1C2t \dfrac { T }{ { T }_{ 0 } } =\dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right)  }{ { C }_{ 1 }{ C }_{ 2 } } t
    or  T=T0exp(KA(C1+C2)t C1C2 )  T={ T }_{ 0 }exp\left( \dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right)t  }{ { C }_{ 1 }{ C }_{ 2 } }  \right) 

  • Question 8
    1 / -0
    A vessel contains a mixture consisting of m1=7kg{m}_{1}=7kg of nitrogen (M1=28)\left( { M }_{ 1 }=28 \right) and m2=11g{m}_{2}=11g of carbon dioixide (M2=44)\left( { M }_{ 2 }=44 \right) at temeprature T=300KT=300K and pressure P0=1atm{ P }_{ 0 }=1\quad atm. The density of the mixture is:
    Solution

    Let V is the volume of the vessel.

    Now, let p1p_{1} and p2p_{2} be the partial pressure, then using gas law: 

    p1V=m1M1RTp_{1}V = \dfrac{m_1}{M_1}RT\\

    p2V=m2M2RT, p0 =p1+ p2p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0}  = p_{1} +  p_{2}\\

    p0=(m1M1+m2M2)RTVp_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\

    V=(m1M1+m2M2)RTp0V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\

    ρmix=(m1+m2)V\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\

    ρmix=(m1+m2)M1M2(m1M2+m2M1)×p0RT\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\

    Substituting values,

    ρmix=(7+11)×28×44×103(7×44+11×28))×1058.3×300\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\

    =1.446 per litre= 1.446 \ per \ litre

    Option A is correct.

  • Question 9
    1 / -0
    When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to  (25)th \\ \\ \left(\dfrac { 2}{ 5 }\right)^{ th } of the distance between the lower and the upper fixed points. The temperature of liquid in K is
    Solution
    Given:
    TbTf=100T_b-T_f=100
    Temperature of liquid, T=25T=\dfrac{2}{5} distance between lower and upper fixed point
    Temperature of liquid, T=25×100=400CT=\dfrac{2}{5}\times 100=40^0C
    On kelvin scale, Tk=273.15+40T_k=273.15+40
                               Tk=313.15KT_k=313.15K
    The correct option is B.
  • Question 10
    1 / -0
    Heat from the sun is received by the earth through
    Solution

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