Question

an electric hater is required to raise the temperature of 90 liters of water from 10*c to 80*c in an hour. if its efficiency is 90%, the cost of operation if energy cost 90 paise/KWh, will be

Answer 1

Answer:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60

o

C

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60

o

C,Q=msΔT

S= specific heat =4200J/kg

o

C

pt=msΔT

2000×t=200×4200×60

t=25200

or t=25.2×10

3

sec.

Answer 2

Answer:

n electric hater is required to raise the temperature of 90 liters of water from 10*c to 80*c in an hour. if its efficiency is 90%, the cost of operation if energy cost 90 paise/KWh, will be