HEAT AND THERMAL ENERGY: Heat Energy Absorbed by a Pot of Water

Q: A 3 kg aluminum pot is filled to capacity with 5 kg of water. If the pot and water are both raised from 25^o C to 95^o C, what total quantity of heat has been absorbed by the system?

A: Although the pot and water both rise and reach a common final temperature, they will not absorb the same quantity of heat. Different substances have a different capacity to manage heat energy, and these differences are summed up with the equation for specific heat capacity ( c ):

c = Heat per unit mass per unit temperature.

c = ( Q / kg*K )

If a substance has a relatively high ( c ) value, it will take more heat energy to raise a given mass of it by each degree Kelvin that it rises, and vice versa. By using known values of ( c ) in regard to aluminum and water, we can determine how much heat was absorbed in the aforementioned scenario:

Note: Please recall that the Celcius and Kelvin scales are divided into increments that are equal in size. For this reason, a change of temperature on one scale directly corresponds to a change of temperature that occurs when measurements are made using the other scale, and vice versa.

c_aluminum = ( 897 J / kg*K )

c_water = ( 4,181 J / kg*K )

Q = mcΔT

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Q_Al = m_Alc_AlΔT_Al

Q_Al = ( 3 kg )( 897 J / kg*K )( 70 K )

Q_Al = 1.88 x 10⁵ J

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Q_H2O = m_H2Oc_H2OΔT_H20

Q_H2O = ( 5 kg )( 4,181 J / kg*K )( 70 K )

Q_H2O = 1.46 x 10⁶ J

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Q_total = Q_Al + Q_H2O

Q_total = 1.88 x 10⁵ J + 1.46 x 10⁶ J

Q_total = 1.65 x 10⁶ J