The temperature of the two outer surfaces of a composite slab consisting of two materials having coefficient of thermal conductivity K and 2K and thickness x and 4x respectively are T₂ and T₁(T₂ > T₁). The rate of heat transfer through the slab, in a steady state is \(\left ( \frac{A(T_{2}-T_{1}K)}{x} \right )\) with fequal to

Solution

A kettle with 2 litre water at 27°C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160J/sec, when its lid is open.In how much time will water heated to 77°C with the lid open? (specific heat of water = 4.2 kJ/kg)

8 min 20 sec
6 min 2sec
14 min
7min

Solution

By the law of conservation of energy, energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.

Pt = ms Δθ+ energy lost

1000t = 2 × 4.2 × 10³ × 50 + 160t

840t = 8.4 × 10³ × 50 = 500 sec = 8min 20 sec

Two marks on a glass rod 10 cm apart are found to in creasetheir distance by 0.08 mm when the rod is heated from 0°C to 100°C. A flask made of the same glass as that of rod measuresa volume of 1000cc at 0°C. The volume it measures at 100°C in cc is

1002.4
1004.2
1006.4
1008.2

Solution

The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167WK^-1m^-1 and thickness1 cm. The temperature is maintained by circulating oil as shown in figure. Find the radiation loss to the surrounding in Jm^-2s^-1 if temperature of the upper surface of the disc is 27°C and temperature of the surrounding is 27°C

595 Jm^-2s^-1
545 Jm^-2s^-1
495 Jm^-2s^-1
None of these

Solution

The rate of heat loss per unit area due to radiation

= ∈σ(T⁴–T₀⁴)

= 0.6 × 5.67 × 10^-8[(400)⁴–(300)⁴] = 595 Jm^–2s^-1
.

Two rods of the same length and areas of cross-section A1and A2have their ends at the same temperature. K₁ and K₂ are the thremal conductivities of the two rods. The rate of flow of heat is same in both rods if

\(\frac{A_{1}}{A_{2}}=\frac{K_{1}}{K_{2}}\)
\(\frac{A_{1}}{A_{2}}=\frac{K_{2}}{K_{1}}\)
A₁A₂= K₁K₂
\(A_{1}K_{1}^{2}=A_{2}K_{2}^{2}\)

Which one of the following graphs best represents the ways in which the total power P radiated by a black body depends upon the thermodynamic temperature T of the body?

Solution

The total power radiated by a black body of area A at temperature TK is given by P = AσT⁴

Where σ= Stefan's constant

= 5.7 × 10^–8 W m^-2 K^–4

Which is best represented in graph. (c)

Three identical rods A, B and C of equal lengths and equal diameters are joined in series as shown in figure. Their thermal conductivities are 2k, k and k/2 respectively.The temperatures at two junction points are

85.7, 57.1°C
80.85, 50.3°C
77.3, 48.3°C
75.8, 49.3°C

The rate of heat flow through the cross-section of the rod shown in figure is (T₂ > T₁ and thermal conductivity of the material of the rod is K)

\(\frac{k\pi r_{1}r_{2}(T_{2}-T_{1})}{L}\)
\(\frac{k\pi (r_{1}+r_{2})^{2}(T_{2}-T_{1})}{4L}\)
\(\frac{k\pi (r_{1}+r_{1})^{2}(T_{2}-T_{1})}{L}\)
\(\frac{k\pi (r_{1}+r_{1})^{2}(T_{2}-T_{1})}{2L}\)

Solution

A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1 :4. If the free face of brass is at 100°C and that of copper at 0°C, the temperature of interface is

80°C
20°C
60°C
40°C

Solution

4K (100 - θ)= K (θ - 0)⇒ 400 - 4θ =θ

⇒ θ = 80°C

Two rods of same length and transfer a given amount of heat 12 second, when they are joined as shown in figure (i).But when they are joined as shwon in figure (ii), then they will transfer same heat in same conditions in

24 s
13 s
15 s
48 s

Solution

\(t\alpha \frac{\iota }{A},t'\alpha \frac{2\iota }{A/2}\)

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete