Tag: heat and modes of heat transfer

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Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A body P is connected to a large body Q through a conducting rod of length. I crossectional  area. A and thermal conductivity K. This assembly is placed in an an atmosplere of temperature ${ T } _{ A }$ and body Q is also maintained at temperature ${ T } _{ A }$. Let beat capacity of body P is C and it is unitally at temperature ${ T } _{ 1 }$. If in time t second temperature of body P falls to ${ T } _{ 2 }$. Then chose the correct option .

  1. $log\left[ \dfrac { { T } _{ 2 }-{ T } _{ A } }{ { T } _{ 1 }-{ T } _{ A } } \right] =\left[ { K } _{ 1 }\dfrac { KA }{ LC } \right] t$

  2. $log\left[ \dfrac { { T } _{ 2 }-{ T } _{ A } }{ { T } _{ 12 }-{ T } _{ A } } \right] =\left[ { K } _{ 1 }\dfrac { KA }{ LC } \right] t$

  3. $log\left[ \dfrac { { T } _{ 1 }-{ T } _{ A } }{ { T } _{ 2 }-{ T } _{ A } } \right] =\left[ { K } _{ 1 }\dfrac { KA }{ LC } .t \right]$

  4. $log\left[ \dfrac { { T } _{ 1 }-{ T } _{ A } }{ { T } _{ 2 }-{ T } _{ A } } \right] =\left[ { K } _{ 1 }\dfrac { KA }{ LC } \right] t$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cooling of body P follows Newton's Law of Cooling, where the rate of heat loss dQ/dt = -C(dT/dt) = (kA/L)(T - Ta). Integrating this leads to the natural log form ln((T1 - Ta)/(T2 - Ta)) = (kA/LC)t.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is $36^o$C. The temperature difference across the layer A is?

  1. $6^o$C

  2. $12^o$C

  3. $18^o$C

  4. $24^o$C
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

a wall has two layers $A$ and $B,$ each made of  a different material.Both the layers have the same thickness.The thermal conductivity of the material of $A$ is twice that of $B.$ Under thermal equilibrium, the temperature difference across the wall is ${36^ \circ }C$ The temperature difference across the layer $A$ is  

  1. ${6^ \circ }C$

  2. ${12^ \circ }C$

  3. ${18^ \circ }C$

  4. ${24^ \circ }C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

An aluminium meter rod of area of cross section $4cm^2$ with K=0.5 cal $g^{-1}$ $^oC^{-1}$ is observed that at steady state 360 cal of heat flows per minute.
The temperature gradient along the rod is

  1. $3^oC/cm$

  2. $6^oC/cm$

  3. $12^oC/cm$

  4. $20^oC/cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat flow H = kA(dT/dx). Given H = 360 cal/min = 6 cal/sec, k = 0.5, A = 4. 6 = 0.5 * 4 * (dT/dx). 6 = 2 * (dT/dx), so dT/dx = 3 C/cm.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A conducting ring lies fixed on a horizontal plane. If a charged nonmagnetic particle is released from a point (on the axis) at some height from the plane, then :

  1. an induced current will flow in clockwise or anticlockwise direction in the loop depending upon the nature of the charge

  2. the acceleration of the particle will decrease as it comes down

  3. the rate of production of heat in the ring will increase as the particle comes down

  4. no heat will be produced in the ring
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A charged particle moving along the axis of a conducting ring does not change the magnetic flux through the ring (the magnetic field lines are parallel to the axis or symmetric). Therefore, no emf is induced and no heat is produced.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Ratio of radius of curvature of cylindrical emitters of same type is $1:4$ and their temp. are in ration $2:1$. Then ration of amount of heat emitted by them is-(For Cylinder length = radius);-

  1. 2:1

  2. 1:1

  3. 4:1

  4. 1:4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power radiated P = sigma * A * T^4. For a cylinder, A = 2*pi*r*L + 2*pi*r^2. If L=r, A = 4*pi*r^2. P is proportional to r^2 * T^4. Ratio P1/P2 = (r1/r2)^2 * (T1/T2)^4 = (1/4)^2 * (2/1)^4 = (1/16) * 16 = 1:1.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

If the coefficient of conductivity of aluminium is $0.5cal/cm-sec-^oC,$ then in order to conduct $10cal/sec-cm^2$ in the steady state, the temperature gradient in aluminium must be

  1. $5^oC/cm$

  2. $10^oC/cm$

  3. $20^oC/cm$

  4. $10.5^oC/cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat flux H/A = k * (dT/dx). Given H/A = 10, k = 0.5. 10 = 0.5 * (dT/dx), so dT/dx = 20 C/cm.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

a rod of length 1 m having cross-sectional area 0.75 $m^{2}$ conduts heat at 6000 $Js^{-1}$. Then the temperature difference across the rod is, if k=200 $Wm^{-1}$ $K^{-1}$

  1. $20^{\circ}C$

  2. $40^{\circ}C$

  3. $80^{\circ}C$

  4. $1000^{\circ}C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using H = kA(dT/L), 6000 = 200 * 0.75 * (dT/1). 6000 = 150 * dT, so dT = 40 C.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A sphere, a cube and a thin circular plate all made of same substance and all have same mass. These are heated to $200^{o}C$ and then placed in a room. Then the:-

  1. Temperature of sphere drops to room temperature at last.

  2. Temperature of cube drops to room temperature at last.

  3. Temperature of thin circular plate drops to room temperature at last.

  4. Temperature of all the three drops to room temperature at the same time
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate of cooling dT/dt = (sigma * A * e * (T^4 - Ta^4)) / (m * c). For same mass and material, the body with the smallest surface area A cools the slowest. A sphere has the smallest surface area for a given volume/mass, so it takes the longest to cool.