# A Brass Boiler Has a Base Area of 0.15 M2 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 - Physics

A brass boiler has a base area of 0.15 m2 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. The thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.

#### Solution 1

Base area of the boiler, = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Mass, m = 6 kg

Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s –1 m–1 K–1

Heat of vaporisation, L = 2256 × 103 J kg–1

The amount of heat flowing into water through the brass base of the boiler is given by:

theta = (KA(T_1-T_2)t)/l  .....(i)

where

T1 = Temperature of the flame in contact with the boiler

T2 = Boiling point of water = 100°C

Heat required for boiling the water:

θ = mL … (ii)

Equating equations (i) and (ii), we get:

:.mL = (KA(T_1-T_2)t)/l

T_1 - T_2 = (mLl)/(KAt)

= (6xx2256xx10^3xx0.01)/(109xx0.15xx60)

= 137.98 ""^@C

Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

#### Solution 2

Here K = 109 Js^(-1) m^(-1) K^(-1)

A  = 0.15 m^2

d = 1.0 cm  = 10^(-2) m

T_2 = 100^@ C

Let T_1 =  temperature of the part of the boiler in contact with the stove.

if Q be the amount of heat flowing per second throught the base of  the boiler, then

Q = (109 xx 0.15 xx (T_1 - 100))/10^(-2) ..(i)

= 1635 (T_1 - 100 )Js^(-1)

Also heat of vaporisation of water L = 2256 xx 10^3 J kg^(-1)

Rate of boiling of water in the boiler, M  = 6.0 kg min^(-1) = 6.0/60 = 0.1 kg^(-1) s

:. Heat received by water per second, Q = ML

=> Q = 0.1 xx 2256 xx 10^3 Js^(-1)  ... ii

:. From equation i and ii we get

1635 (T_1 - 100) = 2256 xx 10^2

or T_1 - 100 = (2256xx10^2)/1653 = 138

T_1 = 138 + 100 = 238^@C

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 11 Thermal Properties of Matter
Q 20 | Page 296