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Exercises · 10.18

Q.A brass boiler has a base area of 0.15 m20.15\ \text{m}^{2} and thickness 1.0 cm1.0\ \text{cm}. It boils water at the rate of 6.0 kg/min6.0\ \text{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 m−1 K−1= 109\ \text{J s}^{-1}\ \text{m}^{-1}\ \text{K}^{-1}; Heat of vaporisation of water =2256×103 J kg−1= 2256 \times 10^{3}\ \text{J kg}^{-1}.

Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★est
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At steady state the heat conducted through the brass base equals the heat used to boil the water: kA(T−100)d=dmdt Lv\dfrac{kA(T-100)}{d} = \dfrac{dm}{dt}\,L_v. Solving for the flame-contact temperature gives T≈238 ∘CT \approx 238\ ^\circ\text{C}.

Why this approach works

When water boils steadily, all the energy that turns it to steam is conducted through the brass base from the hotter flame. At steady state the rate of heat conducted through the metal exactly equals the rate of heat absorbed as latent heat. Equating the two lets us find the flame-side temperature without measuring it directly.

Tflame=Twater+(dm/dt) Lv dk AT_{\text{flame}} = T_{\text{water}} + \frac{(dm/dt)\,L_v\,d}{k\,A}

Step 1 - Heat needed to boil the water per second

dmdt=6.0 kg60 s=0.10 kg s−1\frac{dm}{dt} = \frac{6.0\ \text{kg}}{60\ \text{s}} = 0.10\ \text{kg s}^{-1}

Qt=dmdt Lv=(0.10)(2256×103)=2.256×105 J s−1\frac{Q}{t} = \frac{dm}{dt}\,L_v = (0.10)(2256\times10^{3}) = 2.256\times10^{5}\ \text{J s}^{-1}

Step 2 - Steady-state conduction through the base

With k=109 J s−1m−1K−1k = 109\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}, A=0.15 m2A = 0.15\ \text{m}^2, d=1.0 cm=0.01 md = 1.0\ \text{cm} = 0.01\ \text{m} and Twater=100 ∘CT_{\text{water}} = 100\ ^\circ\text{C}: …

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