Question

· A TKW electric kettle contains 2 litre water
27°C, the kettle is operated for 10 minutes. If
heat is lost to the surroundings at a constant rate
of 160 J/Sec, the temperature attained by the
water in 10 minutes will be ( Sp. heat of water
4.2 KJ/Kg °C)
1) 57°C
2) 67°C
3) 77°C
AJ 87°C​

Answer 1

Heat energy = P × t

P × t = m.s.∆ø

160 × 10 × 60 = 2 × (4.2) × 10³ × (ø2 - ø1 )

=> 10 = (ø2-ø1)

=>10 + 27 = ø2

ø2 = 37°C