Question

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?

Answer 1

Answer:

$3.253\text{ KJ}$

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:

$\begin{gathered} Q\text{ = 79.5 }\times\text{ 0.930 }\times(67-23) \\ Q\text{ = 79.5 }\times0.930\times\text{ 44} \\ Q\text{ = 3,253.14 J} \end{gathered}$

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

Thus, we have the value in KJ as 3.253 KJ