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 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.)

Answer 1

(a) 220.46 Watt

Step-by-step 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