Q. 45.0( 1 Vote )

# Water is boiled in a container having a bottom of surface area 25 cm^{2}, thickness 1.0 mm and thermal conductivity 50W m^{–1} °C^{–1}. 100g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = 2.26 × 10^{6} J kg^{–1}.

Answer :

**Given:**Area of the bottom of container: A = 25 cm

^{2}= 0.0025 m

^{2}.

Thickness of the container: x = 1 mm = 0.001 m

Thermal conductivity of the container: K =50 W m

^{–1}°C

^{–1}.

Latent heat of vaporization : L = 2.26 × 10

^{6}J kg

^{–1}.

Mass of the water converted to steam : m = 100 g = 0.1 kg

Temperature of the water : T

_{2}= 100 ° C

**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.

Also,

Δθ = Q =L× m

Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.

Also, Δt = 1 minute = 60 seconds

Let temperature of the bottom of the container be T

_{1}.

Substituting we get,

∴

∴

∴

Hence, Temperature of the bottom of the container is

130.13 ° C.

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