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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-minute time interval.

(a) Calculate the average power required to warm the water to 100°C in 6.00 minutes. (The specific heat of water is 4186 J/kg·°C.)

Answer :

Final answer:

To find the average power required to warm water from 20°C to 100°C using an immersion heater in 6 minutes, use the formula for power. By plugging in the given values into the formula, you can calculate the average power to be 9,409 W.

Explanation:

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. To calculate the average power required, we will use the formula for power which is P = (m * c * ΔT) / t, where m is the mass of water, c is the specific heat of water, ΔT is the temperature change, and t is the time taken.

Plugging in the values gives us:

P = (237 g * 4186 J/kg · °C * (100°C - 20°C)) / (6 min) = 9,409 W.

Therefore, the average power required to warm the water to 100°C in 6.00 min is 9,409 W.

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Rewritten by : Barada

Answer:

(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