Answer :


Given:
Area of the bottom of container: A = 25 cm2 = 0.0025 m2.
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 × 106 J kg–1.
Mass of the water converted to steam : m = 100 g = 0.1 kg
Temperature of the water : T2 = 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 T1.
Substituting we get,






Hence, Temperature of the bottom of the container is
130.13 ° C.


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