Question

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.

Answer 1

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

Step-by-step explanation:

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

`Q = ms\Delta T`

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

`Q = 40(4186)(80 - 40)`

`Q = 6.69 * 10^6 J`

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

`t = (Q)/(P)`

`t = (6.69 * 10^6)/(20 * 10^3)`

`t = 335 s`