Question

An aluminum electric tea kettle with a mass of 500 g is heated with a 500-W heating coil. How long will it take to heat up 1.0 kg of water from 18°C to 98°C in the tea kettle? The specific heat of aluminum is 900 J/kg ? K and that of water is 4186 J/kg ? K.

Answer 1

heating coil power = 500W = 500 J/sec

initial temperature of tea kettle is not given

case 1) initial temperature of kettle is 20 °C ( 293 K i.e. - room tem )

heat supplied by the coil in time T = WT = 500T

heat required to raise tem of water from 18 °C to 98° C = msT = 1*4186 *80 = 334880 J

heat required to raise tem of kettle20°C to 98° C =   msT = 0.5 *900*60 = 27000 J

heat balance

heat supplied = heat consumed

500T = 334880+27000

T = 723.76 sec = 12.06 min

case 1) initial temperature of kettle is 18C ( 291 K i.e water tem)

heat supplied by the coil in time T = WT = 500T

heat required to raise tem of water from 18 °C to 98° C =  msT = 1*4186 *80 = 334880 J

heat required to raise tem of kettle18°C to 98° C =   msT = 0.5 *900*58 = 26100 J

heat balance

500T = 334880+26100

T = 721.96 sec = 12.03 min