Question

A 300.0 W immersion heater is used to heat 0.250 kg of water from 10.0 °C to 70.0 °C. About how many seconds did this take? (There are 4186 Joules per kcal.)

Answer 1

209.3 seconds

Step-by-step explanation:

P = 300 W, m = 0.250 kg, T1 = 10 degree C, T2 = 70 degree C

c = 4186 J / kg C

Heat given to water = mass x specific heat of water x rise in temperature

H = 0.250 x 4186 x (70 - 10)

H = 62790 J

Power = Heat / Time

Time, t = heat / Power

t = 62790 / 300 = 209.3 seconds