A hot water tank containing 40kg of water is heated by an electric immersion heater rated at 3kilowatt, 240V. Calculate the time required to raise the temperature of the water from 20°C to 60°C assuming that 80% of the heat supplied is retained by the water.

Answer 1

First, we need to calculate the amount of heat required to raise the temperature of the water from 20°C to 60°C.

The specific heat capacity of water is 4.18 Joules/gram°C. Therefore, the specific heat capacity of 40kg (40,000 grams) of water is:

40,000 grams * 4.18 J/g°C = 167,200 J/°C

To calculate the total heat required, we use the formula:

Heat = mass * specific heat capacity * change in temperature

Heat = 40,000 grams * 4.18 J/g°C * (60°C - 20°C)
Heat = 40,000 grams * 4.18 J/g°C * 40°C
Heat = 66,880,000 J

Since only 80% of the heat supplied by the immersion heater is retained by the water, the effective heat supplied is:

Effective Heat = 80% * 66,880,000 J
Effective Heat = 53,504,000 J

Now, we can calculate the time required to supply this amount of heat, using the formula:

Power = Heat / Time

Since the power of the immersion heater is given in kilowatts, we need to convert it to watts:

Power = 3 kW * 1000 = 3000 W

3000 W = 53,504,000 J / Time

Simplifying:

Time = 53,504,000 J / 3000 W
Time ≈ 17833.33 s

Therefore, it would take approximately 17833.33 seconds to raise the temperature of the water from 20°C to 60°C.