a 200 L electric water heater uses 15 kW of electric power. If the herter is 90% efficient, how long does it take to heat the water in the tank from 16 C to 61 C?

Please try to put your class subject in the subj column. "College" is totally meaningless.

time*heatrate*efficiency=200*Cwater*(61-16)
solve for time. Your heat rate is 15kJ/sec, Cwater= 4.8KJ/Kg, and the units on the 200 is kg

check the specific heat of water.

To calculate the time it takes to heat the water in the tank, we can use the formula:

Energy = Power × Time

First, we need to determine the energy required to heat the water in the tank. The formula to calculate the energy required to heat a substance is:

Energy = mass × specific heat capacity × temperature difference

Given:
Mass of water = 200 L = 200 kg (since 1 L of water weighs approximately 1 kg)
Specific heat capacity of water = 4.186 J/g°C (approximate value)
Initial temperature = 16°C
Final temperature = 61°C

Temperature difference = Final temperature - Initial temperature = 61°C - 16°C = 45°C

Energy = 200 kg × 4.186 J/g°C × 45°C = 376,740 J

Next, we need to calculate the time using the formula:

Time = Energy / (Power × Efficiency)

Given:
Power = 15 kW = 15,000 W
Efficiency = 90% = 0.90

Time = 376,740 J / (15,000 W × 0.90) = 27.16 seconds

Therefore, it takes approximately 27.16 seconds to heat the water in the tank from 16°C to 61°C.