The quantity of heat Q that changes the temperature ?T of

Chapter 9, Problem 37TAS

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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 quantity of heat Q that changes the temperature \(\Delta T\) of a mass m of a substance is given by \(Q=c m \Delta T\), where c is the specific heat capacity of the substance. For example, for \(\mathrm{H}_{2} \mathrm{O}, \quad c=1 \quad \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). And for a change of phase, the quantity of heat Q that changes the phase of a mass m is Q = mL, where L is the heat of fusion or heat of vaporization of the substance. For example, for \(\mathrm{H}_{2} \mathrm{O}\), the heat of fusion is 80 cal/g (or 80 kcal/kg) and the heat of vaporization is 540 cal/g (or 540 kcal/kg). Use these relationships to determine the number of calories to change (a) 1 kg of \(0^{\circ} \mathrm{C}\) ice to \(0^{\circ} \mathrm{C}\) ice water, (b) 1 kg of \(0^{\circ} \mathrm{C}\) ice water to 1 kg of \(100^{\circ} \mathrm{C}\) boiling water, (c) 1 kg of \(100^{\circ} \mathrm{C}\) boiling water to 1 kg of \(100^{\circ} \mathrm{C}\) steam, and (d) 1 kg of \(0^{\circ} \mathrm{C}\) ice to 1 kg of \(100^{\circ} \mathrm{C}\) steam.

Text Transcription:

Q = cm Delta T

L_f

L_v

H_2 O, c=1 cal / g cdot ^circ C

H_2 O

0^circ C

100^circ C