Question

The specific heat of aluminum is 0.0215 cal/g°C. If a 4.55 g sample of aluminum absorbs 2.55 cal of energy, by how much will the temperature of the sample change?

Answer 1

The change in temperature is
`26.06^(\circ) C`.

Step-by-step explanation:

It is given that,

The specific heat of Aluminium is cal/g°C

Mass of the sample, m = 4.55 g

Heat absorbed, Q = 2.55 cal

We need to find the change in temperature of the sample. The heat absorbed by an object is given by :

`Q=mc\Delta T`

`\Delta T` is the change in temperature

So,

`\Delta T=(Q)/(mc)\\\\\Delta T=(2.55\ cal)/(4.55\ g* 0.0215 \ cal/g^(\circ) C)\\\\\Delta T=26.06^(\circ) C`

So, the change in temperature is
`26.06^(\circ) C`.