Water is boiled in a container whose bottom has a surface area of 25 cm ^2 , a thickness of 1.0 mm , and a thermal conductivity of 50 W m ⁻¹ ^ C ⁻¹ . After boiling begins, 100 g of water is converted into steam per minute during the steady state. Assuming no heat loss to the atmosphere, determine the temperature of the lower surface of the bottom. The latent heat of vaporization of water is 2.26 10^6 J kg ⁻¹ .

Step-by-step solution

The rate of heat transfer through the bottom of the container is given by Fourier's law of heat conduction: dQ dt = KA(T₂ - T₁) d The rate of heat utilized in boiling the water is: dQ dt = mL t Equating the two rates of heat transfer: KA(T₂ - T₁) d = mL t Substituting the given values, K = 50 W m ⁻¹ ^ C ⁻¹ , A = 25 10⁻⁴ m ^2 , d = 10⁻³ m , m = 0.1 kg , t = 60 s , L = 2.26 10^6 J kg ⁻¹ , and T₁ = 100 ^ C : 50 25 10⁻⁴ (T₂ - 100) 10⁻³ = 0.1 2.26 10^6 60 125(T₂ - 100) = 226000 60 125(T₂ - 100) = 11300 3 T₂ - 100 = 1130