Chapter 15.1, Problem 10SSC

Interpretation Introduction

Interpretation:

The heat absorbed by aluminium is to be calculated when temperature is increased.

Concept introduction:

Specific heat is defined as quantity of heat required to raise the temperature of one gram substance by one Kelvin. It is represented by ‘c’.

Answer to Problem 10SSC

The heat absorbed is $345.3J$

Explanation of Solution

The quantity of heat required can be calculated by using formula:

$\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatabsorbed\\ m=givenmass\\ \Delta T=riseintemperature\\ cisspecificheat\end{array}$

Given data:

$\begin{array}{l}m=5.50g\\ \Delta T=95-25\\ =70°C\\ c=0.897J/g-°C\end{array}$

Substituting the values in formula:

$\begin{array}{c}q=mc\Delta T\\ =5.50\times 0.897\times 70\\ =345.3J\end{array}$

Conclusion

Thus $345.3J$ heat is absorbed to increase the temperature of given amount of aluminium.