Round to the nearest second. A 170−L water heater is rated at 8 kW. If 20 percent of its heat escapes, how long does the heater take to raise the temperature of 170 L of water from 20°C to 60°C?

Answer 1

Hello Nilufar,

Volume f water = 170 L

Mass of water of 170 L = 170 Kg

As 20% of power escapes The effective power = 80% of 8KW = 6.4 KW = 6.4* 10³ J/s

Heat required = M*S* Rise in temperature =Q

M = Mass of water, S = Sp heat capacity of water = 4200 J Rise in temp = 60-20= 40 degree

Power effective = Heat per second

P*Time = Q= 170*4200*40

6.4*10³ * Time = Q

Time ⁼ 170*4200*40/(6.4*10³

This time will be in Seconds ,which needs to be changed in proper unitsw