A tank holding 70 liters of water is heated by a 18000W electric immersion heater. If the specific heat capacity of water is 4200 J/(kg .℃), Calculate the time to change the temperature from 20 Degree Celsius to 100 degree Celsius.

1306.67 s

Explanation:

From the question given above, the following data were obtained:

Volume of water = 70 L

Power (W) = 18000 w

Specific heat capacity (C) = 4200 J/g℃

Initial temperature (T1) = 20 ℃

Final temperature (T2) = 100 ℃

Time (t) =?

Next, we shall determine the mass of the water. This can be obtained by converting 70 L to kg. This is illustrated below:

1 L = 1 kg

Therefore,

70 L = 70 Kg

Thus, the mass of the water is 70 kg.

Next, we shall determine the heat energy involved. This is can be obtained as follow:

Mass (M) of water = 70 Kg

Specific heat capacity (C) = 4200 J/g℃

Initial temperature (T1) = 20 ℃

Final temperature (T2) = 100 ℃

Heat Energy (Q) =?

Q = MC(T2 – T1)

Q = 70 × 4200 × (100 – 20)

Q = 294000 × 80

Q = 3520000 J

Thus, the heat energy involved is 3520000 J

Finally, we shall determine the time taken.

Power (W) = 18000 w

Heat Energy (Q) = 3520000 J

Time (t) =?

Power = energy / time

18000 = 3520000 / time

Cross multiply

18000 × time = 3520000

Divide both side by 18000

Time = 3520000 / 18000

Time = 1306.67 s