Question

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

Answer 1

Step 1: Use the formula of amount of flow of heat.
The amount of heat flowing into water through the brass base of boiler is given by:

$\Delta Q=\frac{KA(T_2-T_1)t}{l}$

Where,

Coefficient of thermal conductivity of brass, $K=109J/smK$

Base area of the boiler, $A=0.15m^2$

Let the temperature of the flame, $T_2$

Inside temperature, $T_1={100}^{\circ }C$ (boiling temperature of water)​

Time gap, $t=1\text{min}=60s$

Thickness of the boiler, $l=1.0\text{cm}=0.01m$

Therefore, Substituting the values we get,

$\Delta Q=\frac{109\times 0.15\times (T_2-100)\times 60}{0.01}J...\left(i\right)$

This heat is used for vaporizing the water
Now,

Bioling rate of water, $R=6.0kg/min$

Mass, $m=6\text{kg}$

Heat of vaporization, $L=2256\times {10}^{3}J/kg$

Step 2: Use latent heat of vaporization formula.

$\Delta Q=mL=6\times 2256\times {10}^3=13536000J...\left(ii\right)$​​​​​​

Step 3: Find the value of temperature of flame.
Equation equation (i) and (ii), we get,

$\frac{109\times 0.15\times (T_2-100)\times 60}{0.01}=13536000$

$\Rightarrow T_2-100=\frac{13536000\times 0.01}{109\times 0.15\times 60}$

$\Rightarrow T_2-100=137.98$

$\Rightarrow T_2-137.98+100={237.98}^{\circ }C$

Therefore, the temperature of the part of the flame in contact with the boiler is ${237.98}^{\circ }C$.