Ask Question
16 November, 12:44

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

+5
Answers (1)
  1. 16 November, 12:52
    0
    (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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Engineering if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers