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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