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

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

$Q = ms\Delta T$

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

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

$Q = 6.69 \times 10^6 J$

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

$t = \frac{Q}{P}$

$t = \frac{6.69 \times 10^6}{20 \times 10^3}$

$t = 335 s$