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What quantity of heat (in kJ) will be absorbed by a 79.5 gpiece of aluminum (specific heat = 0.930 J/g.°C) as itchanges temperature from 23.0°C to 67.0°C?

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Answer:


3.253\text{ KJ}

Step-by-step explanation:

Here, we want to calculate the quantity of heat in KJ

Mathematically, we can calculate that as follows:


Q\text{ = mc}\Delta T

where:

m is the mass of the aluminum piece which is 79.5 g

c is the specific heat capacity of the aluminum which is 0.930 J/g.°C

deLta T is the change in temperature which is the difference in temperature of the final and initial temperature

Substituting the values, we have it that:


\begin{gathered} Q\text{ = 79.5 }*\text{ 0.930 }*(67-23) \\ Q\text{ = 79.5 }*0.930*\text{ 44} \\ Q\text{ = 3,253.14 J} \end{gathered}

To convert this to KJ, we have to divide by 1000

Thus, we have the value in KJ as 3.253 KJ

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