Water is boiled in a container having a bottom of surface area 25 cm2, thickness 1.0 mm and thermal conductivity 50 W m−1°C−1. 100 g 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 vaporisation of water = 2.26 × 106 J kg−1.
Answers
The temperature of the lower surface of the bottom is T = 130°C
Explanation:
Area of the bottom of the container "A" = 25 cm^2 = 25 × 10^−4 m^2
Thickness of the bottom of the container, l = 1 mm = 10−-3 m
Latent heat of vaporisation of water, L=2.26×106 J kg−1 L=2.26×106 J kg-1
Thermal conductivity of the container, k=50 W m−1°C−1 k=50 W m-1°C-1
Mass "m" = 100 g = 0.1 kg
Solution:
Rate of heat transfer is given by :
ΔQ / Δt = mL / Δt = (0.1) × 2.26 × 10^6 / 1 min
ΔQ / Δt = 0.376 × 10^4 J/s
ΔQ / Δt = ΔT / l ÷ kA
0.376 × 10^4 = T − 100 / 10^−3 / 50 × 25 × 10^−4
0.376 × 10^4 = 50 × 25 × 10^−4 (T − 100) / 10^−3
(T − 100) = 3.008 × 10
T = 130°C
Hence the temperature of the lower surface of the bottom is T = 130°C
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