How long (in seconds) will it take to heat 40 kg of water from 40°C to 80°C with a 20-kW immersion heater? Assume no energy is lost to the environment and specific heat of water is 4.186 kJ/kgoC.

Respuesta :

Answer:

Total time taken by the heater to raise the temperature of water is 335 s

Explanation:

As we know that the heat required to raise the temperature of water is given as

[tex]Q = ms\Delta T[/tex]

here we know that

m = 40 kg

s = 4186 J/kg C

now its temperature is raised from 40 degree to 80 degree

so we will have

[tex]Q = 40(4186)(80 - 40)[/tex]

[tex]Q = 6.69 \times 10^6 J[/tex]

Now we know that the power of the heater is 20 kW

so total time taken by the heater to raise the temperature of water is given as

[tex]t = \frac{Q}{P}[/tex]

[tex]t = \frac{6.69 \times 10^6}{20 \times 10^3}[/tex]

[tex]t = 335 s[/tex]