Q. 13.8( 4 Votes )

# A uniform slab of dimension 10 cm × 10 cm × 1 cm is kept between two heat reservoirs at temperatures 10°C and 90°C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W m^{–1} °C^{–1}. Find the amount of heat flowing through the slab per minute.

Answer :

**Given:**Area of the uniform slab: A = 10× 10 cm

^{2}= 100 cm

^{2}= 0.01 m

^{2}

Height of the slab : x = 1 cm = 0.01 m

Temperature difference of two heat reservoirs:

ΔT = 90-10=80° C

The thermal conductivity of the material: K=0.80 W m

^{–1}°C

^{–1}.

**Formula used:**

Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred , ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.

Substituting we get,

Hence, the amount of heat flowing through slab is 3840 J/min.

Rate this question :

Assume that the total surface area of a human body is 1.6 m^{2} and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant σ is 6.0 × 10^{−8} W m^{−2} K^{−4}.

Calculate the amount of heat radiated per second by a body of surface area 12 cm^{2} kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface = 0.80 and σ = 6.0 × 10^{−8} W m^{−2} K^{−4}.

Four identical rods AB, CD, CF and DE are joined as shown in figure. The length cross-sectional area and thermal conductivity of each rod are ℓ, A and K respectively. The ends A, E and F are maintained at temperatures T_{1}, T_{2} and T_{3} respectively. Assuming no loss of heat to the atmosphere, final the temperature at B.

HC Verma - Concepts of Physics Part 2

Find the rate of heat flow through a cross-section of the rod shown in figure (θ_{2}> θ_{1}). Thermal conductivity of the material of the rod is K.

HC Verma - Concepts of Physics Part 2