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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

It takes \(880 \mathrm{J}\) to raise the temperature of \(350 \mathrm{g}\) of lead from 0 to \(20.0^{\circ} \mathrm{C} .\) What is the specific heat of lead?
A 75 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) breaks off from a glacier, slides along the frictionless ice to the ground from a height of $2.43 \mathrm{m},$ and then slides along a horizontal surface consisting of gravel and dirt. Find how much of the mass of the ice is melted by the friction with the rough surface, assuming \(75 \%\) of the internal energy generated is used to heat the ice.
A stainless steel saucepan, with a base that is made of \(0.350-\mathrm{cm}-\) thick steel \([\kappa=46.0 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})]\) fused to a \(0.150-\mathrm{cm}\) thickness of copper $[\kappa=401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})],\( sits on a ceramic heating element at \)104.00^{\circ} \mathrm{C} .\( The diameter of the pan is \)18.0 \mathrm{cm}$ and it contains boiling water at \(100.00^{\circ} \mathrm{C} .\) (a) If the copper-clad bottom is touching the heat source, what is the temperature at the copper-steel interface? (b) At what rate will the water evaporate from the pan?
A chamber with a fixed volume of \(1.0 \mathrm{m}^{3}\) contains a monatomic gas at \(3.00 \times 10^{2} \mathrm{K} .\) The chamber is heated to a temperature of \(4.00 \times 10^{2} \mathrm{K}\). This operation requires \(10.0 \mathrm{J}\) of heat. (Assume all the energy is transferred to the gas.) How many gas molecules are in the chamber?
A 2.0 -kg block of copper at \(100.0^{\circ} \mathrm{C}\) is placed into $1.0 \mathrm{kg}$ of water in a 2.0 -kg iron pot. The water and the iron pot are at \(25.0^{\circ} \mathrm{C}\) just before the copper block is placed into the pot. What is the final temperature of the water, assuming negligible heat flow to the environment?
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