Question

an electric heater of power rating 3000W of water equivalent 200g contains 800g of water. what is the time required to raise the temperature of water from 20C to 80C

Answer 1

Answer:

Heat required=(40×10

3

)×1×(80−15)=26×10

6

cal

Now, H=

R×4.2

V

2

t

cal

or, t=

V

2

HR×4.2

∴t=

220×220

26×10

5

×12×4.2

=2708 seconds.

Answer 2

Given:

Power, P = 3000 W

Water equivalent, w = 200 gm = 0.2 kg

Mass of water, m = 800 gm = 0.8 kg

Initial temperature, T1 = 20°C

Final temperature, T2 = 80°C

To Find:

The time required to raise the temperature.

Calculation:

- Specific heat of water, c = 4200 J/kg°C

- The heat required is given as:

H = P × t    ...(i)

- Heat can also be calculated as :

H = (m + w) × c × ΔT   ...(ii)

- From (i) and (ii), we have:

P × t = (m + w) × c × ΔT

⇒ 3000 × t = (0.8 + 0.2) × 4200 × (80 - 20)

⇒ t = 1 × 4200 × 60/3000

⇒ t = 42 × 2

⇒ t = 84 s

- So, the time required to raise the temperature from 20°C to 80°C is 84 seconds.