400 g of water are emptied at 20 ° C in a vessel. The container is placed at

Fire of a flame that gives it a certain amount of thermal energy of
So that it takes 5 minutes for the water to reach a temperature of 90 ° C.
If the vessel does not absorb thermal energy, a) how much energy did the
Water ?, b) what power does the system heat the water? Answer
The same questions assuming that the container and the environment absorb the
50% of the energy delivered by the thermal source.

I need help, I transformed cal/m to watts but it gives 390.761 Watts, and the exercise says it is 3,33 watts. Can somebody guide me in how to make this exercise. (All except "a")

Thanks for sharing your answer. It is right. I obtained 390.5 watts.

To solve this exercise, we need to use the specific heat formula and calculate the energy absorbed by the water, and then find the power using the given time.

a) To calculate the energy absorbed by the water, we use the specific heat formula:

Q = m * c * ΔT

Where:
Q is the energy absorbed or released (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).

Given:
m = 400g
Initial temperature, T1 = 20°C
Final temperature, T2 = 90°C

The specific heat capacity of water is approximately 4.18 J/g°C.

Using the formula, we can calculate the energy absorbed:

Q = 400g * 4.18 J/g°C * (90°C - 20°C)
Q = 400g * 4.18 J/g°C * 70°C
Q = 117080 J

Therefore, the energy absorbed by the water is 117080 Joules.

b) To find the power, we need to know the time taken for the water to reach the final temperature. In this case, the time is given as 5 minutes. However, we need to convert minutes to seconds as power is calculated in joules per second (watts).

1 minute = 60 seconds

So, 5 minutes = 5 * 60 = 300 seconds.

Now we can calculate the power:

Power = Energy / Time

Power = 117080 J / 300 s
Power = 390.27 Watts

For the second part of the exercise, assuming that the container and the environment absorb 50% of the energy delivered by the thermal source, we need to calculate the revised energy and power.

a) The energy absorbed by the water is reduced by 50% due to absorption by the container and the environment:

Energy_absorbed = 117080 J * 0.5
Energy_absorbed = 58540 J

Therefore, the energy absorbed by the water in this case is 58540 Joules.

b) To calculate the revised power, we use the same formula:

Power = Energy / Time

Power = 58540 J / 300 s
Power = 195.13 Watts

Hence, the power in this case is 195.13 Watts.

It seems there was an error in converting calories to watts. The correct value for the power when the container and the environment absorb 50% of the energy delivered by the thermal source is 195.13 Watts, not 3.33 Watts. It appears there may have been a calculation mistake in converting the units.

Please double-check your calculations to ensure accuracy.