An industrial electric boiler rated at 4 kW contains 612 kg of water at 80°C. To adjust boiler level to safe value, 78 kg of water at 26°C was added. Determine the: (i) new temperature of water in the boiler; (ii) time required to restore water temperature to 80°C. (7 marks)craft1 electrical June/July 2020

i. To solve this problem, we can use the formula for the specific heat capacity of water:

c = Q / (m * ΔT)

where c is the specific heat capacity of water, Q is the heat added to the water, m is the mass of the water, and ΔT is the change in temperature.

We can rearrange this formula to solve for Q:

Q = c * m * ΔT

First, we need to calculate the heat added to the water when it was added to the boiler. We can do this by using the specific heat capacity of water, which is 4.186 J/g°C:

Q = 4.186 J/g°C * 78 kg * (80°C – 26°C) = 214,054 J

Next, we need to calculate the new temperature of the water in the boiler. We can do this by using the formula for the heat capacity of a substance, which is given by:

C = Q / ΔT

where C is the heat capacity of the substance, Q is the heat added to the substance, and ΔT is the change in temperature.

We can rearrange this formula to solve for ΔT:

ΔT = Q / C

The heat capacity of the water in the boiler is given by:

C = c * m = 4.186 J/g°C * (612 kg + 78 kg) = 3,003,108 J/°C

Substituting these values into the formula for ΔT, we get:

ΔT = 214,054 J / 3,003,108 J/°C = 0.07°C

The new temperature of the water in the boiler is 80°C + 0.07°C = 80.07°C

ii. Finally, we need to calculate the time required to restore the water temperature to 80°C. We can do this by using the formula:

t = Q / (P * η)

where t is the time required, Q is the heat added to the water, P is the power of the heating element, and η is the efficiency of the heating element.

Since the boiler is rated at 4 kW, the power of the heating element is 4 kW. The efficiency of the heating element is not given, so we will assume it is 100% efficient (η = 1).

Substituting these values into the formula, we get:

t = 214,054 J / (4 kW * 1) = 53,513 s

The time required to restore the water temperature to 80°C is approximately 53,513 seconds, or about 14.9 hours.