Solved: The specific heat capacity of ice is about 0.5

Chapter 9, Problem 38TAS

(choose chapter or problem)

The quantity of heat with temperature change is \(Q=c m \Delta T\). For a change of phase of water it is Q = mL, where \(L_{f}\) is the heat of fusion, 80 cal/g, and \(L_{v}\) is the heat of vaporization, 540 cal/g.

The specific heat capacity of ice is about \(0.5 \quad \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and suppose that it remains at that value all the way to absolute zero. Calculate the number of calories it would take to change a 1-g ice cube at absolute zero (\(-273^{\circ} \mathrm{C}\)) to 1 g of boiling water. How does this number of calories compare with the number of calories required to change the same gram of \(100^{\circ} \mathrm{C}\) boiling water to \(100^{\circ} \mathrm{C}\) steam?

Text Transcription:

Q = cm Delta T

L_f

L_v

0.5 cal / g cdot ^circ C

-273^circ C

100^circ C