Chapter 15, Problem 76A

How much heat is absorbed by a 44.7-g piece of leadwhen its temperature increases by 65.4°C?

Interpretation Introduction

Interpretation:

The energy required to heat a given amount of lead is to be calculated.

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 76A

The heat absorbed is$377.1J$

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:

$\begin{array}{l}m=44.7g\\ \Delta T=65.4°C\\ c=0.129J/g-°C\end{array}$

Substituting the values in formula:

$\begin{array}{c}q=mc\Delta T\\ =44.7\times 0.129\times 65.4\\ =377.1J\\ =10.74kJ\end{array}$

Conclusion

Thus$377.1J$heat is required to raise the temperature of given amount of lead.