As the table is part of parameters of a drinking water machine nameplate, the internal circuit is simplified as shown in the figure. When S is closed, the S1 is opened, the water dispenser in the insulating state, when the S and S1 closed at the same time, the drinking water machine in the heating state. [c water =4.2 * l03J/kg - C]Rated voltage 220VRated heating 1000WThermal insulation 44WThe capacity of the water heater 2L(1) the current in the normal working state of the drinking water heater(2) the electric energy consumed by the water cooler to keep 50min(3) if the heat loss is ignored, the dispenser will heat 2kg water from 30 to 90 C. How long will it take?
(3) the volume of 2L water:V=2L=2dm3=2 * 10-3m3,According to P =MVAvailable water quality:M= P V=1.0 * 103kg/m3 * 10-3m3=2kg * 2,The heat absorbed by water:Q=cm t=4.2 * 103J/ (kg C) * 2kg * (90 -30 OC) =5.04 * 105J,Regardless of heat loss, time consuming energy is equal to the amount absorbed by water,
(1) from the table data, the power of the drinking machine under the thermal insulation condition is 44W,According to P=UI, the current at normal operation:I=PU=44W220V=0.2A;(2) the electric energy consumed by the water heater for insulation and 50min:W=Pt=44W x 50 x 60s=1.32 * 105J;
According to W=Pt, heating time:T1=QP heating=5.04 x 105 J1000W=504s.Answer: (1) when the water heater is in thermal insulation, the current is 0.2A when working normally;(2) the electric energy consumed by the water heater 50min is 1.32 * 105J;(3) if the heat loss is ignored, the water dispenser heats the water of 2L from 30 DEG C to 90 DEG C and requires 504s.
