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.)

(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

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)