1.5 kg of liquid water initially at 12∘C is to be heated to 95∘C in a teapot equipped with an 800-W electric heating element inside. The specific heat of water can be taken to be 4.18 kJ/kgK, and the heat loss from the water during heating can be neglected. The time it takes to heat water to the desired temperature is

(a) 5.9 min
(b) 7.3 min
(c) 10.8 min
(d) 14.0 min
(e) 17.0 min

(c) 10.8 min

Explanation:

The energy  that must be supplied by the heating element is W = Pxt, where is the power in watts multiplied by the time. By the energy conservation law, the energy is the heat, q, absorbed by  the water to raise its temperarure by a change in temperature ΔT, and can be calculated from the equation:

q = mcΔT

ΔT = 95 ºC - 12 ºC = 83 ºC = 83 K ( because ºC is linearly related to K)

q = 1.5 kg x 4.18 kJ/kg x (83 K) = 520.4 kJ = 5.2 x 10⁵ J

Thus

W = P x t ⇒ t = W/P

t = 5.2 x 10⁵ J / 800 J/s = 650 s x 1 min/ 60s = 10.8 min

answer:

true

explanation:

yes indeed

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