Thermal Physics: Important Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

1) The outer ends of two bars A & B (inner ends of which are joined together by welding) are at 100°C and 50°C respectively. Calculate the temperature at the welded joint if they have the same cross‒section and the same length and their thermal conductivities are in the ratio of A:B = 7:5.

Give data

K₁: K₂ = 7:5;

d₁ = d₂;

θ₁ = 100°C;

θ₃ = 50°C

Solution

Formula:

Q/t  = K₁/d_{1 .} A(θ₁‒ θ₂) = K₂/d_{2 .} A(θ₂‒ θ₃)

K₁ / K₂ = (θ₂‒ θ₃) / (θ₁‒ θ₂)

7/5  = (θ₂‒50) / (100‒θ₂)

i.e.. 700 ‒ 7θ₂ = 5θ₂ ‒ 250

12 θ₂ = 950

θ₂ = 79.166°C

Temperature at the welded joint 79.166°C

2) A composite metal bar of uniform cross‒section is made up of 0.25m of metal A and 0.1 m of metal B and each being in perfect thermal contact with the adjoining part. There is no heat loss at the sides. The thermal conductivities of metals A & B are 920 and 140 S.I units respectively. The end A is maintained at 100°C and the end B is maintained at 24°C. Calculate the temperature at A‒B junction.

Given data

K₁ = 920;

K₂ = 140;

d₁ = 0.25m;

d₂ = 0.10m;

θ₁=100°C

θ₃ = 24°C

Solution

Formula:

K₁ / K₂ = d₁/d₂ . (θ₂‒ θ₃)/(θ₁‒ θ₂)

35θ₂‒840 = 9200 ‒ 92θ₂

127θ₂ = 10040

 θ₂ = 79.05°C

∴ Temperature at A‒B junction = 79.05°C

3) Two bars of copper and steel of length 1.0m and 0.5m respectively and of co‒efficient of thermal conductivity 400 W/m‒K and 50W/m‒K respectively are jointed end to end. The free ends of copper and steel are maintained at 100°C ad 0°C respectively. Calculate the temperature of copper‒steel junction if both bars have the same area of cross‒section.

Given data

K₁ = 400 W/m‒K;

K₂ = 50 W/m‒K;

d₁ = 1m ;

d₂ = 0.5m;

θ₁ = 100°C

θ₃ = 0°C

Solution

Formula:

K₁ / K₂ = d₁/d₂ . (θ₂‒ θ₃) / (θ₁‒ θ₂)

 θ₂ / (100‒ θ₂) = 4

i.e.,  400 ‒ 4θ₂ = θ₂

5θ₂ = 400

θ₂ = 80°C

The temperature of Copper ‒ Steel junction = 80°C

4) A copper rod of length 50 cm and cross‒sectional area 6 ×10^‒2 cm² is connected in series with an iron rod of same area of cross‒section and length 25 cm. One end of copper is immersed in boiling water. The far end of the iron rod is in an ice bath of 0°C. Find the rate of transfer of heat from boiling water to ice bath. (Thermal conductivity of copper and iron are 401 Wm^‒1K^‒1 and 80 Wm^‒1K^‒1 respectively).

Given data

d₁ = 50 × 10^‒2 m

θ₁ = 100°C = 373 K

K₁ = 401 Wm^‒1 K^‒1

d₂ = 25 × 10^‒2 m

θ₂ = 0° C = 273K

K₂ = 80 Wm^‒1K^‒1

Area A = 0.06 x 10^‒4 m²

Solution

Formula:

Rate of heat transfer (or) Amount of heat conducted per second Q =

∴ Q = 6 × 10^‒4 / 43.71×10^‒4

Q = 0.1372 Watts

∴ Rate of transfer of heat = 0.1372 Watts.

ADDITIONAL SOLVED PROBLEMS

1) A 30 cm length of iron rod is heated at one end to 100°C, while the other end is kept at a temperature of 35°C. The area of cross section of the iron rod is 0.725 cm². Assume that the iron rod is thermally insulated. Calculate the amount of heat conducted through the rod in 8 minutes along the way. Given the thermal conductivity of iron K = 62 Wm^‒1K^‒1

Solution

The quantity of heat conducted

Q = KA(θ₁‒ θ₂)t / x

 = [ 62 × 0.725 × 10^‒4 x (373‒308) × 480 ] / 0.3

 = 467.48 J

The iron rod conducts 467.48 joules in 8 minutes.

2) The total area of a glass window pane is 0.8m². Calculate how much heat is conducted per hour through the glass window pane if thickness of glass is 3mm. The temperature of the inside surface is 25°C and of the outside surface is 4°C. Thermal conductivity of glass is 1.1 Wm^‒1K^‒1_.

Solution

Amount of heat conducted = Q = KA(θ₁‒ θ₂)t / x

 = [ 1.1x0.8×21 × 3600 ] / 3×10^‒3

 =  2.217 × 10⁷ Joules.

Heat conducted per hour through the window pane = 2.217×10⁷ Joules.

3) The total area of the glass window pane is 0.9m². Calculate how much heat is conducted per hour through the glass window pane if thickness of the glass 4mm, the temperature of the inside is 29°C and of the outside surface is 3°C. Thermal conductivity of glass is 1.1 Wm^‒1K^‒1.

Solution

Amount of heat conducted

Q = KA(θ₁‒ θ₂)t / x

 = [1.1×0.9 × 26 × 3600] / 4×10^‒3

 = 2.316 × 10⁷ Joules.

Heat conducted per hour = 2.316 × 10⁷ Joules

4) Calculate the thickness of the slab of area 90×10^‒4 m² through which 6 Joules of heat is flowing per second through the opposite faces maintained at a temperature difference of 20K. The coefficient of thermal conductivity of the material of the slab is 0.04 Wm^‒1K^‒1

Solution

Amount of heat conducted = Q = KA(θ₁‒ θ₂)t / x

Thickness x = KA(θ₁‒ θ₂)t / Q

 x = [ 0.04 × 90×10^‒4 × 20×1 ] / 6

The thickness of the slab = 1.2 × 10^‒3 m

5) A glass vessel with an insulating cover and with a surface area of 75cm² and 12mm thick is filled with ice at 0°C and placed in a second vessel maintained at a temperature of 90°C. Find the mass of the ice that melts per minute when the flow of heat becomes steady. Specific latent heat of ice = 3.36×10⁵ J kg^‒1K^‒1 and for glass = 1.1 Wk^‒1m^‒1.

Solution

Amount of heat conducted = Q = KA(θ₁‒ θ₂)t / x

Q = [ 1.1× 75×10^‒4 × 90×60 ] / 12×10^‒3

= 3712.5 J

If M kg of ice melts, the heat required to melt ice i.e. M×L must be equal to Q

 M×3.36×10⁵ = 3712.5

Mass of the ice that melts per minute (M) = 0.01104 kg (or) 11.04gm

6) A solid square of side 50 cm and thickness 10 cm is in contact with steam at 100°C on one side. A block of ice at 0°C rests on the other side of the solid. 5 kg. of ice is melted in one hour. Calculate the thermal conductivity of the solid.

Given:

Specific latent of ice = 3.36×10⁵ J Kg^‒1K^‒1

Solution

Amount of heat conducted = Q = KA(θ₁‒ θ₂)t / x

Q = [ K(50×10 × 10^‒4) × 100 × 3600 ] / 10×10^‒2

Q = 180000 K

During one hour 180000K of heat is conducted through the solid square and which is responsible for melting of 5 kg. of ice

 M×L = Q

 5×3.36×10⁵ = 180000K

K = ( 5×3.36×10⁵) / (180000)

Thermal Conductivity of the solid = 9.33 Wm^‒1K^‒1

ASSIGNMENT PROBLEMS

1) A 40cm length of iron rod is heated at one end to 90°C while the other end is kept at a temperature of 30°C. The area of cross section of the iron rod is 0.7 cm². Assume that the iron rod is thermally insulated. Calculate the amount of heat conducted through the rod in 6 minutes along the way. Given that the thermal conductivity of iron is K= 62 Wm^‒1K^‒1          (Ans: 234.36J)

2) Calculate the thickness of the slab of area 85 × 10^‒4 m² through which 8 joules of heat is flowing through the opposite faces maintained at a temperature difference of 30K. The coefficient of thermal conductivity of the material of the slab is 0.05 Wm^‒1K^‒1. The time taken for the heat flow is 10 seconds          (Ans : 0.0159m)

3) Equal bars of copper and aluminium are welded end to end. If the free ends of copper and aluminium are maintained at 150°C and 0°C respectively. Find the temperature of the welded interface. Assume the thermal conductivity of copper and aluminium to be 386.4 Wm^‒1K^‒1 and 210 Wm^‒1K^‒1 respectively.         (Ans. 97.183°C)

4) A slab consists of two parallel layers of different materials 4cm and 2cm thick and of thermal conductivity 226.8 Wm^‒1K^‒1 and 150.2 Wm^‒1K^‒1 respectively. If the opposite faces of the slab are at 100°C and 0°C. Calculate the temperature of the surface dividing the two.          (Ans. 315.857K)