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Thermal Properties of Matter Test - 44

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Thermal Properties of Matter Test - 44
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  • Question 1
    1 / -0
    The thermal conductivity of a material in CGS system is 0.4. In steady state the rate of flow of heat $$10 \,cal/s-cm^2$$, then the thermal gradient will be
    Solution
    $$\dfrac{\Delta Q}{\Delta t}=\dfrac{KA\Delta T}{\Delta x}$$
    Thermal gradient $$\dfrac{\Delta T}{\Delta x}$$ is given by 
    $$\dfrac{\left ( \dfrac{\Delta Q}{\Delta t} \right )}{KA}=\dfrac {10}{0.4}= 25^o C/ cm$$
  • Question 2
    1 / -0
    A building at a temperature T(in k) is heated by an ideal heat pump which uses the amosphere at $$T_0(k)$$. What is the equilibrium temperature of the building ? 
  • Question 3
    1 / -0
    Two metal rods, rod-1 and rod-2 of equal length are welded together end to end.Under steady state conditions, when the free end of the end of the rod-1 is kept at $$100^0C$$  and the free end of the rod -2 is kept at $$0^0C$$ , the temperature at the welded junction is approximately.
    [Thermal conductivity of rod-1 = $$92  J s^{-1}m^{-1}K^{-1}$$, and rod-2 =$$16 J s^{-1}m^{-1}K^{-1}$$]
    Solution
    Heat Current = $$i = \dfrac{KA \Delta T}{L}$$
    $$K_1(T_1-T) = K_2(T-T_2)$$
    $$92(100-T)=16(T-0)$$
    $$\implies T = 85^o C$$
  • Question 4
    1 / -0
    The radii of two spheres made of same metal are $$r$$ and $$2r$$. These are heated to the same temperature and placed in the same surrounding. The ratio of rates of decrease of their temperature will be:
    Solution
    Same metal $$\implies$$ same density
    Rate of cooling
    $$\dfrac { -dT }{ dt } =\dfrac { eA\sigma  }{ mc } \left( { T }^{ 4 }-{ T }_{ 0 }^{ 4 } \right) $$ formula used.
    T is same for both spheres
    $${ T }_{ 0 }$$ is same
    L is same
    $$\sigma $$
    e is same
    Here rate of cooling proportional $$\propto \dfrac { A }{ m } $$
    $$\propto \dfrac { 4{ \pi r }^{ 2 } }{ D\times 4{ \pi r }^{ 3 } } $$
    Density  $$\propto\dfrac { 1 }{ Dr } $$, D=Density =Mass/volume
    $${ R }_{ 1 }$$ = rate of cooling of radius $${ r }_{ 1 }$$
    $${ R }_{ 2 }$$ = rate of cooling of radius $${ r }_{ 2 }$$
    $$\dfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\dfrac { { D }_{ 2 }{ \quad r }_{ 2 } }{ { D }_{ 1 }\quad { r }_{ 2 } } \quad \left[ { D }_{ 1 }={ D }_{ 2 }\quad same\quad density \right] $$
    $$=2:1$$

  • Question 5
    1 / -0
    'Thermos Flask' was invented by
    Solution

  • Question 6
    1 / -0
    Thermal radiation can be detected by.
    Solution
    Thermal radiation can be detected by thermometer, radiometer, bolometer.
  • Question 7
    1 / -0
    Which one of the following statements is not true regarding thermal radiations?
    Solution
    All bodies emit thermal radiation at all temperatures.They are electromagnetic in nature and have same speed. But they cannot be reflected by mirror as thermal energy is not visible.
  • Question 8
    1 / -0
    A cube, a thin circular plate, and a sphere are made of the same metal and having the same mass and same volume. They are initially heated to the same temperature. When left in air at the room temperature, which of these bodies will cool fastest and slowest respectively?
    Solution

    For a given volume, among all the three objects, the sphere has the least surface area and the cube will have the maximum surface area. This is also due to the fact that the volume and mass of all three are the same.

    According to Heat theory, the rate of loss of heat is proportional to the surface area of the object. Hence, the rate of cooling will be slowest for the sphere and fastest for the cube.

    Thus the correct option is A.

  • Question 9
    1 / -0
    The temperature (T) of one mole of an ideal gas varies with its volume (V) as $$T = -\alpha { V }^{ 3 }+\ \beta { V }^{ 2 }$$, where $$\alpha$$ and $$\beta$$ are positive constants. The maximum pressure of gas during this process is
    Solution

    $$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$$

    For maximum pressure

    $$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$$

     

  • Question 10
    1 / -0
    Two identical glass bulbs are interconnected by a thin glass tube at $$0^{\circ}C$$. A gas is filled at N.T.P. in these bulb is placed in ice and another bulb is placed in hot bath, then the pressure of the gas becomes $$1.5\ times$$. The temperature of hot bath will be

    Solution
    Quantity of gas in these bulbs is constant i.e,
    Initial no. of moles in both bulb $$=$$ final no. of moles
    $$\Rightarrow n_1+n_2={ n }_{ 1 }^{ ' }+{ n }_{ 2 }^{ ' }$$
    $$\Rightarrow \dfrac{PV}{R(273)}+\dfrac{PV}{R(273)}=\dfrac{1.5PV}{R(273)}+\dfrac{1.5PV}{R(T)}$$
    $$\Rightarrow \dfrac{2}{273}=\dfrac{1.5}{273}+\dfrac{1.5}{T}$$
    $$\Rightarrow T=819K$$
              $$=546°C$$
    Hence, the answer is $$546°C.$$
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