The heating curve for a sample of pure ethanol is provided above. The temperature was recorded as a \(\pu{50.0 g}\) sample of solid ethanol was heated at a constant rate. Which of the following explains why the slope of segment \(T\) is greater than the slope of segement \(R\) ?

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When heat is added but the temperature remains constant, the sample would be in the process of changing phases.

\(T\) corresponds to the region when ethanol is a gas. \(R\) refers to the region when ethanol is a liquid.

The heat corresponds to temperature with the following equation:

$$ q = mc\Delta T $$

If constant heat is being applied but the temperature is increasing at a greater rate, it must mean that the \(c\) value (specific heat) is different for the the liquid and gas phases.

$$ q=mc\Delta T $$ $$ \frac{q}{m}=c\Delta T $$ $$ k = c\Delta T \text{ where } \frac{q}{m} \text{ is a constant}$$

If mass and heat are constant, the specific heat and temperature change are inversely related. Therefore a greater temperature change corresponds to a lower specific heat value and vice-versa.