Here, base area A = 0.15 m2

Here, T = temperature of the part of the flame in Contact With boiler. Rate of boiling of water = 6 kg min-1. Rate at Which heat is being Supplied to water is:

= 2.2556 X 105 Js-1

Equating the above two values of Δθ/Δt
1635 (T - 1000) = 2.256 X 105
⇒ T - 100 = (2.256 X 105)/ 1635 = 138
∴ T = 138 + 100 = 238 °C.