Question

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

Answer 1

In this question, we need to determine the temperature of the part of the flame in contact with the boiler such that the base area and the thickness of the brass boiler are 0.15${m^2}$ and 1.0 cm respectively. For this, we will use the relation between the heat losses per unit time, the base area of the boiler, the thickness of the boiler, the thermal conductivity of the material and boiling rate of water.

Complete step by step answer:
The base area of the boiler, $A = 0.15{m^2}$
The thickness of the boiler, $l = 1.0cm = 0.01m$
Boiling rate of water, $R = 6.0{\text{ kg/min}}$
So we can say the mass of the water $m = 6kg$
Time is taken $t = 1\min = 60\sec$
Thermal conductivity of brass$K = 109J{s^{ - 1}}{m^{ - 1}}{K^{ - 1}}$
The heat of vaporization of water$L = 2256 \times {10^3}Jk{g^{ - 1}}$
We know that the heat loss per unit time is given by the formula
$\dfrac{{dQ}}{{dt}} = \dfrac{{kA\Delta T}}{L} - - (i)$
We can also write equation (i) as
$Q = \dfrac{{kA\Delta T}}{L}t - - (ii)$
Now we substitute the values in equation (ii), so we get
$Q = \dfrac{{109 \times 0.15 \times \left( {T - 100} \right)}}{{0.01}} \times 60 - - (iii)$
We also know that the heat loss is given by the formula
$Q = mL - - (iv)$
Hence by substituting the values in equation (iv), we get

$$Q = mL \\\ \Rightarrow Q= 6 \times 2256 \times {10^3} - - (v) \\\$$

Now equate equation (iii) with equation (v) to find the temperature
$\dfrac{{109 \times 0.15 \times \left( {T - 100} \right)}}{{0.01}} \times 60 = 6 \times 2256 \times {10^3}$
Hence by solving

$$\dfrac{{109 \times 0.15 \times \left( {T - 100} \right)}}{{0.01}} \times 60 = 6 \times 2256 \times {10^3} \\\ \Rightarrow T - 100 = \dfrac{{6 \times 2256 \times 1000 \times 0.01}}{{60 \times 109 \times 0.15}} \\\ \Rightarrow T - 100 = 158 \\\ \therefore T = {258^ \circ }C \\\$$

Therefore, the temperature of the part of the flame in contact with the boiler is ${258^ \circ }C$.

Note: It is to be noted here that, all the units of the measuring data should be converted into either SI units or any similar units before substituting them in the associated formula.