Numerical 1

A bar 0.2 m in length and 2.5 cm² in cross section is ideally lagged. One end is maintained at 100^oC and the other end is maintained at 0^oC by immersing in melting ice. Calculate the mass of ice melt in one hour. Thermal conductivity of the material of the bar is 4 x 10² Wm^-1K^-1.

Solution

Length of bar (l) = 0.2 m

Cross section area (A) = 2.5 cm² = 2.5 x 10^-4 m²

Temperature of one end [latex](\theta_1)[/latex] = 100^oC

Temperature of other end [latex](\theta_2)[/latex] = 0^oC

Mass of ice melted (m) =?

Time (t) = 1 hour = 60 x 60 = 3600 s

Thermal conductivity of the material of the bar (K) = 4 x 10² Wm^-1K^-1

We know that,

Q₁ = [latex]\frac{KA(\theta_1 – \theta_2)t}{l}[/latex] …………………. (i)

Also, quantity of heat required to melt ice

Q₂ = mL_f ……………. (ii)

According to question,

Q₁ = Q₂

Or, [latex]\frac{KA(\theta_1 – \theta_2)}{l}t = mL_f[/latex]

Or, m = [latex]\frac{KA(\theta_1 – \theta_2)}{lL_f}t[/latex]

      = [latex]\frac{(4\times 10^2)\times (2.5\times 10^{-4})\times (100-0)}{0.2\times 3.36\times 10^5}\times 3600[/latex]           

= 0.536 kg.

Numerical 2

A pot with a steel bottom 8.5 mm thick rest on a hot stove. The area of the bottom of the pot is 0.15 m². The water inside the pot is at 100^oC and 390 gm of water is evaporated every 3 minutes. Find the temperature of lower surface of the pot which is in contact with the stove. (Thermal conductivity of pot = 50.2 Wm^-1K^-1, latent heat of vaporization = 2256 x 10³ J/Kg).

Solution

Thickness of steel bottom (x) = 8.5 mm = 8.5 x 10^-3 m

Area of pot (A) = 0.15 m²

Temperature of pot inside [latex](\theta_2)[/latex] = 100^oC = 100 + 273 = 373 K

Mass of water evaporated (m) = 390 gm = 0.39 kg

Time (t) = 3 minutes = 3 x 60 = 180 s

Temperature of lower surface of the pot [latex](\theta_1)[/latex] =?

We know that,

Q₁ = [latex]\frac{KA(\theta_1 – \theta_2)}{x}t[/latex] …………………. (i)

Also, quantity of heat required to vaporize water

Q₂ = mL_v ……………. (ii)

According to question,

Q₁ = Q₂

Or, [latex]\frac{KA(\theta_1 – \theta_2)}{x}t = mL_v[/latex]

Or, [latex](\theta_1-\theta_2) = \frac{mL_vx}{KAt}[/latex]

Or, [latex](\theta_1 – 373) = \frac{0.39\times (2256\times 10^3)\times (8.5\times 10^{-3})}{50.2\times 0.15\times 180}[/latex]

Or, [latex]\theta = 5.518 + 373[/latex]

      = 378.518 K

      = 378.518 – 273

     = 105.518^oC.

Numerical 10

Estimate the rate of heat loss through a glass window of area 2 m² and thickness 4 mm when the temperature of the room is 300 K and that of air outside is 5^oC. (Thermal Conductivity of glass = 1.2 Wm^-1K^-1).

Solution

Rate of heat loss (Q/t) =?

Area of window (A) = 2 m²

Thickness (x) = 4 mm = 4 x 10^-3 m

Temperature of room [latex](\theta_1)[/latex] = 300 K

Temperature of outside [latex](\theta_2)[/latex] = 5^oC = 5 + 273 = 278 K

We know that,

Q = [latex]\frac{KA(\theta_1 – \theta_2)}{x}t[/latex]

Or, [latex]\frac{Q}{t} = \frac{KA(\theta_1 – \theta_2)}{x}[/latex]

      = [latex]\frac{1.2\times 2\times (300 – 278)}{(4\times 10^{-3})}[/latex]

      = 13200 Wm^-1K^-1.

Numerical 12

The sun is a black body of surface temperature about 6000 K. If the sun’s radius is 7 x 10⁸ m, calculate the energy per second radiated from its surface. The earth is about 1.5 x 10¹¹ m from the Sun. Assuming all the radiation from the Sun falls on the surface of sphere of this radius, estimate the energy per second per meter square received by the earth. (Stefan’s Constant = 5.7 x 10^-8 Wm^-2K^-4)

Solution

Temperature of sun’s surface (T) = 6000 K

Radius of Sun (R) = 7 x 10⁸ m

Energy per second radiated from its surface (P) =?

We know that,

P = [latex]\sigma AT^4[/latex]

            = [latex]\sigma 4\pi R^2T^4[/latex]

            = (5.7 x 10^-8) x 4 x 3.14 x (7 x 10⁸)² x (6000)⁴

            = 4.546 x 10²⁶ Watt.

Now, [latex]\frac{P}{A’} = \frac{4.546\times 10^{26}}{4\pi r^2}[/latex]

      = [latex]\frac{4.546\times 10^{26}}{4\times 3.14\times (1.5\times 10^{11})^2}[/latex]

            = 1609 Wm^-2.

Numerical 13

A sphere of radius 2.00 cm with a black surface is cooled and then suspended in a large evacuated enclosure with black walls maintained at 27^oC. If the rate of change of thermal energy of sphere is 1.85 Js^-1 when its temperature is – 73^oC, calculate the value of Stefan’s constant.

Solution

Radius of sphere (r) = 2 cm = 0.02 m

Initial temperature (T₁) = 27^oC = 27 + 273 = 300 K

Rate of change of thermal energy (P) = 1.85 Js^-1

Final temperature (T₂) = – 73^oC = – 73 + 273 = 200 K

Stefan’s constant [latex](\sigma)[/latex] =?

We know that,

P = [latex]\sigma A(T_1^4 – T_2^4)[/latex]

Or, [latex]\sigma = \frac{P}{A(T_1^4 – T_2^4)}[/latex]

      = [latex]\frac{P}{4\pi r^2(T_1^4 – T_2^4)}[/latex]

      = [latex]\frac{1.85}{4\times 3.14\times (0.02)^2\times (300^4 – 200^4)}[/latex]

            = 5.665 x 10^-8 Wm^-2K^-4.

Numerical 14

Estimate the power loss through unit area from a perfectly black body at 327^oC to the surrounding environment at 27^oC. (Stefan’s Constant = 5.67 x 10^-8 Wm^-2K^-4).

Solution

Power loss (P) =?

Area (A) = 1 m²

Initial temperature (T₁) = 327^oC = 327 + 273 = 600 K

Final temperature (T₂) = 27^oC = 27 + 273 = 300 K

We know that,

P = [latex]\sigma A(T_1^4 – T_2^4)[/latex]

      = [latex](5.7\times 10^{-8})\times 1\times (600^4 – 300^4)[/latex]
            = 6889.05 Watt.

Numerical 15

A spherical blackbody of radius 5 cm has its temperature 127^oC and its emissivity is 0.6. Calculate its radiant power. (Stefan’s Constant = 5.67 x 10^-8 Wm^-2K^-4).

Solution

Radius of black body (r) = 5 cm = 0.05 m

Temperature (T) = 127^oC = 127 + 273 = 400 K

Emissivity (e) = 0.6

Radiant power (P) =?

We know that,

Radiant power (P) = [latex]\sigma eAT^4[/latex]

      = [latex]\sigma e4\pi r^2T^4[/latex]

      = (5.67 x 10^-8) x 0.6 x 4 x 3.14 x (0.05)² x (400)⁴

      = 27.34 Watt.