Chemistry

Chemistry

Problem

Equal 50.0 g50.0\,\text{g} samples undergo the following changes at constant pressure: Process 1: Liquid water cools from 25 °C25\,\text{°C} to 5 °C5\,\text{°C} (c=4.18 J g−1 °C−1c = 4.18\,\text{J}\,\text{g}^{-1}\,\text{°C}^{-1}). Process 2: Liquid ethanol warms from 5 °C5\,\text{°C} to 25 °C25\,\text{°C} (c=2.44 J g−1 °C−1c = 2.44\,\text{J}\,\text{g}^{-1}\,\text{°C}^{-1}). Process 3: Steam at 100 °C100\,\text{°C} condenses to liquid at 100 °C100\,\text{°C} (ΔHvap=40.7 kJ mol−1\Delta H_{vap} = 40.7\,\text{kJ}\,\text{mol}^{-1}). Rank the three processes from greatest energy released to greatest energy absorbed.
Intuitive explanation

The key is recognizing that sensible heat (temperature change) and latent heat (phase change) operate on completely different scales. For processes 1 and 2, you're moving heat by changing temperature—use q=mcΔTq = mc\Delta T and compare magnitudes. For process 3, a phase change releases or absorbs energy based on molar enthalpy of vaporization, which is typically much larger than sensible heating over a modest temperature range. Calculate each energy value, paying attention to signs (released = negative, absorbed = positive), then order them from most negative to most positive.

Intuitive explanation

The key is recognizing that sensible heat (temperature change) and latent heat (phase change) operate on completely different scales. For processes 1 and 2, you're moving heat by changing temperature—use q=mcΔTq = mc\Delta T and compare magnitudes. For process 3, a phase change releases or absorbs energy based on molar enthalpy of vaporization, which is typically much larger than sensible heating over a modest temperature range. Calculate each energy value, paying attention to signs (released = negative, absorbed = positive), then order them from most negative to most positive.

Browse the problem library