TS EAMCET 19-July-2022 Shift 1 Solved Paper

Given, base area of cooking pot,
A=0.2m2
Thickness, l=2‌cm=2×10−2m
Boiling rate, R=‌

m

t

=3‌kg‌/‌min=‌

1

20

‌kg∕ s
Thermal conductivity, K=120Js−1m−1K−1
Heat of evaporation, L=2×106J/‌kg
The amount of heat flowing into water through the base of cooking pot,
Q=‌

KA(T1−T2)t

l

where, T1= Temperature of the cooking p and
T2= Boiling point of water.
Heat required to boil water, Q′=mL
According to given situation, Q=Q′
‌

KA(T1−T2)

l

t=mL
⇒‌‌T1−T2‌=‌

mLl

KAt

=(‌

m

t

)‌

Ll

KA

‌=‌

1

20

×‌

2×106×2×10−2

120×0.2

=83∘C