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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.

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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`

 

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APPEARS IN

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