A rod of length l with thermally insulated lateral surface, consists of a material whose heat conductivity coefficient varies with temperature as k = a T , where a is a constant. The ends of the rod are kept at temperatures T 1 and T 2 . Find the function T x , where x is the distance from the end whose temperature is T 1 .

Consider a rod of length l and area of cross-section A , whose faces are maintained at temperatures θ 1 and θ 2 , respectively. The curved surface of the rod is kept insulated from the surrounding to avoid the leakage of heat. In steady-state, the heat current, i.e., the rate of heat flowing from one face to the other face is given by, d Q d t = k A d T d x Where, k is the coefficient of thermal conductivity of the material of the rod. Let us consider an element of length d x , at a distance x from the hot end, where the temperature drops from T to T - d T . So, d Q d t = - k A d T d x = - a T A d T d x ⇒ d T T = - 1 a A d Q d t d x On integrating, we get, ∫ T 1 T d T T = - 1 a A d Q d t ∫ 0 x d x ⇒ ln T T 1 = - 1 a A d Q d t x ⇒ T = T 1 e - 1 a A d Q d t x Again, T 2 = T 1 e - 1 a A d Q d t l Thus, ln T T 1 ln T 2 T 1 = - 1 a A d Q d t x - 1 a A d Q d t l = x l ⇒ ln T T 1 = x l ln T 2 T 1 ⇒ T T 1 = T 2 T 1 x l ⇒ T = T 1 T 2 T 1 x l

HARD

JEE Advanced

IMPORTANT

Earn 100

Ice at $0 ° C$ is added to $200 g$ of water, initially at $70 ° C$ in a vacuum flask. When $50 g$ of ice has been added and has all melted, the temperature of the flask and contents is $40 ° C$ . When a further $80 g$ of ice has been added and has all melted, the temperature of the whole becomes $10 ° C$ . Find the latent heat of fusion of ice.

Important Questions on Calorimetry and Heat Transfer

MEDIUM

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 3

A copper cube of mass

200 g

slides down a rough inclined plane of inclination

37 °

at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy, find the increase in temperature of the block as it slides down through

60 cm

. Specific heat capacity of copper is equal to

420 J {kg}^{- 1} K^{- 1}

. (Take

g = 10 m s^{- 2}

)

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 6

An electric heater is placed inside a room of total wall area

137 m^{2}

to maintain the temperature inside at

20 ° C

. The outside temperature is

- 10 ° C

. The walls are made of three composite materials. The innermost layer is made of wood of thickness

2.5 cm

, the middle layer is of cement of thickness

1 cm

and the exterior layer is of brick of thickness

2.5 cm

. Find the power of electric heater, assuming that there are no heat losses through the floor and the ceiling. The thermal conductivities of wood, cement and brick are

0.125 W m^{- 1} ° C^{- 1}, 1.5 W m^{- 1} ° C^{- 1}

and

1.0 W m^{- 1} ° C^{- 1}

, respectively.

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 7

2 m

long wire of resistance

4 Ω

and diameter

0.64 mm

is coated with plastic insulation of thickness

0.06 mm

. A current of

5 A

flows through the wire. Find the temperature difference across the insulation in the steady-state. Thermal conductivity of plastic is

0.16 \times 10^{- 2} cal s^{- 1} {cm}^{- 1} ° C^{- 1}

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 8

Two chunks of metal with heat capacities

C_{1}

and

C_{2}

are interconnected by a rod of length

l

and cross-sectional area

A

and fairly low conductivity

k

. The whole system is thermally insulated from the environment. At a moment

t = 0

, the temperature difference between the two chunks of metal equals

(Δ T)_{0}

. Assuming the heat capacity of the rod to be negligible, find the temperature difference between the chunks as a function of time.

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 9

A rod of length

l

with thermally insulated lateral surface, consists of a material whose heat conductivity coefficient varies with temperature as

k = \frac{a}{T}

, where

a

is a constant. The ends of the rod are kept at temperatures

T_{1}

and

T_{2}

. Find the function

T (x)

, where

x

is the distance from the end whose temperature is

T_{1}

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 10

One end of a uniform brass rod

20 cm

long and

10 {cm}^{2}

cross-sectional area, is kept at

100 ° C

. The other end is in perfect thermal contact with another rod of identical cross-section and length

10 cm

. The free end of this rod is kept in melting ice and when the steady-state has been reached, it is found that

360 g

of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is

0.25 cal s^{- 1} {cm}^{- 1} ° C^{- 1}

and

L = 80 cal g^{- 1}

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 11

Heat flows radially outward through a spherical shell of outside radius

R_{2}

and inner radius

R_{1}

The temperature of the inner surface of the shell is

θ_{1}

and that of outer is

θ_{2}

. At what radial distance from the centre of the shell, the temperature is just half-way between

θ_{1}

and

θ_{2}

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 6 - Calorimetry and Heat Transfer>Exercises>Q 12

A layer of ice of thickness $y$ is on the surface of a lake. The air is at a constant temperature $- θ ° C$ and the ice water interface is at $0 ° C$ . Show that the rate at which the thickness increases is given by

$\frac{d y}{d t} = \frac{K θ}{L ρ y}$

where, $K$ is the thermal conductivity of the ice, $L$ is the latent heat of fusion and $ρ$ is the density of ice.

Important Questions on Calorimetry and Heat Transfer

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