Question

What is the specific heat of the solid phase?

If you could just explain to me how I get it, it'd be great

Answer 1

You can use Q=mc(deltaT) for the first part of the plot.

Answer 2

would the answer be .1 then?

Answer 3

The graph shows three states of the material.

Solid, Liquid, and Gas.

Each phase has a different specific heat.

For the solid phase, the left hand most slope, we can find the specific heat as\[Q = mc \Delta T \]\[c = {Q \over m \Delta T}\]\[c = {200 \over 20 (50)}\]

When finding the specific heat of the other two phases, ignore the horizontal regions.

Answer 4

oh so .2 alright got it thank you :)

Answer 5

Let's not forget our unit though. \[c \left [ J \over g ^{\circ} C \right ]\]

Answer 6

.2 J/g*C :)

Answer 7

The horizontal regions represent heat addition to change states.

These are commonly referred to as "heat of fusion" for melting and "heat of vaporization" for evaporating.

Recall that these phase changes occur without a change in temperature. Also remember that these heats of phase changes are separate from specific heat.

Answer 8

@eashmore can I ask you one more thing?

Answer 9

Sure.

Answer 10

how do I calculate the Latent Heat of Vaporization using the same graph

Answer 11

In a very similar fashion.

Recall that heat of vaporization is the heat required to evaporate a substance at a constant temperature. We use the following equation for the heat required to vaporize a given mass of a substance. \[\large Q_V = mh_v\]where \(\large h_v\) is the heat of vaporization and has units of J/g.

From the chart, we realize that in most cases, matter evaporates at a high temperature. Therefore, vaporization occurs at the second horizontal line at a temperature of 200C.

Note that we add 400 J of heat [1000-600].

We can now solve the above equation for \(h_v\).