Chapter U5.6, Problem 8E

(a)

Interpretation Introduction

Interpretation:The energy needed to transfer in order to raise the temperature of 150 g of aluminum from $20°\text{C}$ to its melting pointneeds to be determined.

Concept Introduction : The formula to calculate heat transfer is,

  $H=mc\Delta T$

Where, H is heat transfer, m is mass, c is specific heat capacity and $\Delta T$ is change of temperature.

Answer to Problem 8E

Heat transferred is $\text{20128}\text{.5}\text{cal}$ .

Explanation of Solution

Specific heat of aluminum is $0.\text{21}\frac{\text{cal}}{\text{g}°\text{C}}$ , Melting point of aluminum is $659°\text{C}$ and mass of aluminum is 150 g.

  $\text{Change}\text{in}\text{temperature (}\Delta T\text{)}\text{=}\text{659}-\text{20}\text{=639}°\text{C}$

Putting all the above values in the equation of heat transfer

  $\begin{array}{c}H=mc\Delta T\\ =150\text{g}\times \text{0}\text{.21}\frac{\text{cal}}{\text{g}°\text{C}}\times \text{639}°\text{C}\\ \text{=20128}\text{.5}\text{cal}\end{array}$

(b)

Interpretation Introduction

Interpretation:The energy needed to transfer in order to melt 150 g of aluminum needs to be determined.

Concept Introduction : The formula to calculate heat transfer using the mass and heat of fusion is,

  $H=mL$

Where, H is heat transfer and L is heat of fusion.

Answer to Problem 8E

The energy needed to transfer in order to melt 150 g of aluminum is 14175 cal.

Explanation of Solution

Heat of fusion of aluminum (L) is 94.5 cal/ g and mass of aluminum is 150 g.

Use the formula to calculate heat.

  $\begin{array}{c}H=mL\\ =150\text{g}\times \text{94}\text{.5}\text{cal/}\text{g}\\ \text{=14175}\text{cal}\end{array}$