A 0.4 kilogram sample of Aluminum at 115 degrees Celsius is put into a container containing 0.5 kilograms of water at 15 degrees Celsius. Neglecting the small amount of energy absorbed by the container and knowing that the specific heat of Aluminum is 900 kJ/kg*C, and the specific heat of the water is 4186 kJ/kg*C answer the following question.

Compared to the heat liberated by the aluminum, the heat absorbed by the water is

greater.

smaller.

the same

what heat the alumin gives off has to be equal to the heat the water gained, No you have an issue here, putting the hot aluminum into dwater, some of the water will vaporize, but if you configure the container so that no heat is lost, that steam will recondense and the heat (and water ) will not be lost.

To compare the heat liberated by the aluminum to the heat absorbed by the water, we will use the principle of conservation of energy, also known as the "heat transfer equation". We can find the heat transferred using the formula:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

First, let's find the heat liberated by the aluminum:

Q_aluminum = (m_aluminum) * (c_aluminum) * (ΔT_aluminum)

Q_aluminum = (0.4 kg) * (900 kJ/kg°C) * (115°C - 15°C)

Q_aluminum = (0.4 kg) * (900 kJ/kg°C) * (100°C)

Q_aluminum = 36,000 kJ

Now, let's find the heat absorbed by the water:

Q_water = (m_water) * (c_water) * (ΔT_water)

Q_water = (0.5 kg) * (4186 kJ/kg°C) * (115°C - 15°C)

Q_water = (0.5 kg) * (4186 kJ/kg°C) * (100°C)

Q_water = 209,300 kJ

Comparing the two values, we can see that the heat absorbed by the water (209,300 kJ) is greater than the heat liberated by the aluminum (36,000 kJ).

Therefore, the correct answer is: The heat absorbed by the water is greater than the heat liberated by the aluminum.

To determine the heat absorbed by the water, we can use the equation:

Q = m * c * ΔT

where:
Q is the heat absorbed or liberated
m is the mass of the substance
c is the specific heat
ΔT is the change in temperature

For the aluminum:
m = 0.4 kg
c = 900 kJ/kg*C
ΔT = final temperature - initial temperature = 115°C - initial temperature

For the water:
m = 0.5 kg
c = 4186 kJ/kg*C
ΔT = final temperature - initial temperature = final temperature - 15°C

Since the small amount of energy absorbed by the container is neglected, we can assume that the final temperature will be the same for both the aluminum and water.

Comparing the two equations, we see that the heat absorbed by the water is determined by the product of its mass, specific heat, and ΔT, just like for the aluminum. Therefore, the heat absorbed by the water is the same as the heat liberated by the aluminum.

So, the answer is: the same.