Hola buenos dias me peuden ayudar

1. Suppose you have an electric heater in your house with a capacity of 110 liters of water that works with a current of 20 A for 140 V. The heater can heat water from 15 C to 50 C in 1 hour 36 minutes. Let's say your family is made up of four people and uses approximately 220 liters of hot water to bathe for bathing daily, at a cost of $2.80 per kilowatt hour.

1.1. What is the electrical power of the heater in kW

1.2. What is the energy spent by the heater in half an hour (write the result in kWh)?

Consider power to be the energy used per unit of time, i.e. P = E/t; where, if power is measured in kW and time in hours, then the energy will be measured in kW h.

1.3. Graph the result of the energy expended by the heater and explain it.

1.4. With the result of the energy expenditure obtained, calculate the cost that your family pays daily for bathing, considering that they spend 220 liters of water; for this:

1.4.1. Calculate the cost of leaving the heater running at its maximum capacity for one hour.

1.4.2. Calculate how much they pay to heat 220 litres of hot water from 15 C to 50 C.

1.5. If the resistance of the heater breaks down, what value must the resistance have to work with a capacity of 220 liters?

1. Now, suppose you have an electric heater shaped like a straight circular cylinder, with a radius of 26.5 cm and a height of 50 cm, which has a capacity of 110 liters of water.

2.1. Calculate the hydrostatic pressure at the bottom of the electric heater when it is full of water. Write down your result in kPa.

2.2. The temperature regulator of the water heater stopped working, so the water inside the water evaporated completely, which caused the heater to fill with steam. The heater contained liquid water halfway, before it started working. So what will be the density of the steam inside the heater? Consider that 1 liter of water has a mass equal to 1 kg.

1. Analyze and respond:

3.1 What law applies in the operation of the heater and why?

3.2 What law was applied to repair the heater and why?