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Chapter 14: Problem 45

Compute the heat of fusion of a substance from these data: \(31.15 \mathrm{kJ}\) will change \(0.500 \mathrm{kg}\) of the solid at \(21^{\circ} \mathrm{C}\) to liquid at \(327^{\circ} \mathrm{C},\) the melting point. The specific heat of the solid is \(0.129 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})\).

Short Answer

Expert verified
Answer: The heat of fusion for the given substance is approximately 22.922 kJ/kg.

Step by step solution

01

Calculate the change in temperature

First, we need to calculate the change in temperature (\(\Delta T\)). This can be done by subtracting the initial temperature from the final temperature: \(\Delta T = T_{final} - T_{initial}\). \(\Delta T = 327^{\circ} \mathrm{C} - 21^{\circ} \mathrm{C} = 306 \mathrm{K}\)
02

Calculate the heat required for the temperature change (Qt)

Using the mass (\(m\)), the specific heat of the solid (\(c\)), and the change in temperature (\(\Delta T\)), we can calculate the heat required for the temperature change (Qt): \(Qt = mc\Delta T\) \(Qt = 0.500 \mathrm{kg} \times 0.129 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K}) \times 306 \mathrm{K} = 19.689 \mathrm{kJ}\)
03

Calculate the heat of fusion (Lf)

Now we have all the information needed to compute the heat of fusion (Lf). We can rearrange the equation derived in the analysis: \(Lf = \frac{Q - Qt}{m}\) Plug in the given values: \(Lf = \frac{31.15 \mathrm{kJ} - 19.689 \mathrm{kJ}}{0.500 \mathrm{kg}} = 22.922 \mathrm{kJ/kg}\)
04

Report the heat of fusion

The heat of fusion for this substance is approximately \(22.922 \mathrm{kJ/kg}\).

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Most popular questions from this chapter

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