Question 18 of 20intermediate🔧 ApplyNumerical4 marks

A brass boiler has a base area of 0.15 m2^2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s−1^{-1} m−1^{-1} K−1^{-1}; Heat of vaporisation of water = 2256×1032256 \times 10^3 J kg−1^{-1}.

Correct Answer

The temperature of the flame in contact with the boiler is approximately 238°C.

Exercise: EXERCISES | Q: 10.18 | (Page 24)
For More Understanding

Explanation

This problem applies the concept of heat conduction through the brass base of the boiler. In steady state, the rate of heat conducted through the brass equals the rate of heat used to vaporize water. The formula for heat conduction is Q/t = KA(T₁ - T₂)/d, where K is thermal conductivity, A is area, d is thickness, and T₁, T₂ are temperatures of hot and cold ends respectively. The boiling point of water (100°C) is taken as T₂, and the heat required for vaporization is calculated using the latent heat of vaporization.

Solution Steps

  1. Step 1: Calculate the rate of heat required to boil water Mass of water boiled per second = 6.0 kg/min ÷ 60 = 0.1 kg/s Rate of heat transfer (Power) = Mass rate × Latent heat P = 0.1×22560.1 \times 2256 × 10³ = 225600 J/s

  2. Step 2: Apply the heat conduction formula In steady state: Heat conducted = Heat used for vaporization P = KA(T₁ - T₂)/d Where: K = 109 J s⁻¹ m⁻¹ K⁻¹, A = 0.15 m², d = 0.01 m, T₂ = 100°C (boiling point of water)

  3. Step 3: Solve for T₁ 225600 = (109×0.15109 \times 0.15 × (T₁ - 100))/0.01 225600 = 1635 × (T₁ - 100) T₁ - 100 = 225600/1635 = 137.98 T₁ = 100 + 137.98 ≈\approx 238°C

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