Question

.It's a cool day, about 10 ^oC, so you plan to make about 5.0 kg of clear soup using your slow cooking crockpot. To decide whether the soup will be ready for dinner, you estimate how long it will take before the soup gets to its boiling point. Before adding the ingredients, you turn the crockpot over and read that it is a 200-ohm device that operates at 120 volts. Since your soup is mostly water, you assume it has the same thermal properties as water, so its specific heat capacity is 4200 J/(kg ^oC) and its heat of vaporization is 2.3 x 10⁶ J/kg.

Answer 1

Let time taken by the soup to reach the boiling point be t seconds

Heat delivered by electrical source in t seconds = Power * Time = (V^2)/R*t = (120*120/200*t) = 72t

Heat required by the soup = m*H* (T2-T1) = 5*4200*(100-10) = 1.89 * 10^6

Here m is the mass of soup and H is specific heat capacity of water. T2 is final temperature and T1 is initial temperature. T2 = 100 degree Celsius which is boiling point of water.

Equating the two heats we get t = (1.89*10^6)/72 = 26250 seconds = 7.29 hours