TS EAMCET 20-July-2022 Shift 2 Solved Paper

Given, density of water, ρw=1000‌kg∕m3
Specific heat capacity of water,
sw=400Jkg−1K−1
Volume of water, V=31=3×10−3m3
∴ Mass of water, m=V×ρw
‌=3×10−3×1000=3‌kg
Change of temperature of water,
∆T‌=[(273+80)−(273+0)]K
‌=80K
Heat required, Q‌=msw∆T
‌=3×42.00×80=1008000
‌=1.008×106J
Heat produced by heater,
H=‌

V2

R

t
⇒1.008×106=‌

(200)2

50

×t
⇒t=1260 s=‌

1260

60

‌min=21‌min