Question

1. the heat of vaporization of ethyl alcohol ( l_{v} ) is about 200 cal/g. that means that 1 gram of this refrigerant upon vaporization would release

a) less than 200 calories.
b) 200 calories.
c) more than 200 calories.

Explanation:

Step1: Understand heat of vaporization

The heat of vaporization \(L_v\) is the amount of heat required for a substance to change from liquid to gas phase. For ethyl alcohol, \(L_v = 200\ cal/g\). When 1 gram of ethyl alcohol vaporizes, it absorbs heat. But the question is about the heat released. Wait, no, actually, when a substance vaporizes, it absorbs heat. But maybe there is a mis - wording. Wait, no, the heat of vaporization is defined as the heat absorbed per unit mass during vaporization. However, if we consider the reverse process (condensation), the heat released per unit mass is the same as the heat of vaporization. But the problem says "upon vaporization". Wait, no, vaporization is a process of liquid to gas, which is an endothermic process (absorbs heat). But maybe it's a translation error. Wait, no, if we assume the problem is correct as per the given data (maybe a wrong - term usage). The heat of vaporization \(L_v\) is the energy per unit mass for vaporization. If we consider the energy associated with the phase change, for 1g, the energy involved in the phase change (vaporization) is \(Q = mL_v\). If we assume that the problem is using "release" in a wrong sense (maybe should be absorb, but if we go by the given options and the definition of \(L_v\) as \(200\ cal/g\) for the phase change (vaporization), the amount of heat for the phase change of 1g is \(200\ cal\).

Answer:

b) 200 calories.