Question

an electric immersion heater has a power rating of 1500 W. If the heater is placed in a liter of water @ a room temperature, how many minutes will it take to bring water to a boil? (assume that there is no heat loss)

Answer 1

let t minutes be required
so Heat energy given by Heater = 1500*t*60 J
and so on..

Answer 2

Heat energy required =Mass*specific heat *change in temperature..

Answer 3

why 60, is it for seconds ?

Answer 4

yup

Answer 5

you need to first find the mass of 1 ltr of water in kg using density of water..

Answer 6

Heat energy given by Heater = 1500*t*60 J

Heat energy required =1000g (1)(20)

Answer 7

mass should be in Kg and not in gms and density should be in Kg/m^3 and

Answer 8

1kg = 1L

Answer 9

we're using grams instead of kg.

Answer 10

not necessarily depends on density...
you can do it in gms and cm but you need to include J (mechanical equivalent of heat energy i.e conversion between calorie and Joules)

Answer 11

where will i put the answer in heat energy required?

Answer 12

so, i'll use 1kg?

Answer 13

Heat is a form of energy, and therefore is measured in joules. There are other units of heat, though; the most common one is the kilocalorie:
One kilocalorie (kcal) is defined as the amount of heat needed to raise the temperature of 1 kg of water by 1 C° (from 14.5°C to 15.5°C).

Answer 14

okayy

Answer 15

room temp = 20 deg.cent.
boiling point = 100 deg.cent.
delta in temp rise = 80 deg.cent

Answer 16

yeah...

Answer 17

energy needed to boil the water = 1kg*4186*80 degrees = 4186*80 = 334880
energy given by the heater to boil the water = 1500*t seconds
t = 334880 / 1500 = 223.2533 seconds = 3.7 minutes

Answer 18

1 kCal = 4186 joules

Answer 19

does the above answer look right?

Answer 20

why 80 degrees?

Answer 21

isn't that the rise in tempreture needed to heat the water from room temp 20 degress to 100 degrees boling point

Answer 22

ahhh,,yeah,,gets

Answer 23

[mc DeltaT\] 1000grams (1)(80),