Ask Question
17 November, 00:29

An office worker uses an immersion heater to warm 237 g of water in a light, covered, insulated cup from 20°C to 100°C in 6.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume that the wire is at 100°C throughout the 6.00-min time interval. (a) Calculate the average power required to warm the water to 100°C in 6.00 min. (The specific heat of water is 4186 J/kg · °C.)

+5
Answers (1)
  1. 17 November, 03:37
    0
    (a) 220.46 Watt

    Explanation:

    m = 237 g

    T1 = 20 degree C, T2 = 100 degree c, t = 6 minutes = 6 x 60 = 360 seconds

    V = 120 V, c = 4186 J/kg C

    (a)

    Heat required to raise the temperature = m x c x (T2 - T1)

    H = 0.237 x 4186 x (100 - 20)

    H = 79366.56 Joule

    Power = Heat / time = 79366.56 / 360 = 220.46 Watt
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “An office worker uses an immersion heater to warm 237 g of water in a light, covered, insulated cup from 20°C to 100°C in 6.00 minutes. The ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers