Chapter 15, Problem 74A

How many joules of heat are lost by 3580 kg of granite asit cools from 41.2°C to -12.9°C? The specific heat ofgranite is $0.803J/(g・°C)$ .

Interpretation Introduction

Interpretation:

The energy lost by granite is to be calculated when heat is lost by a given amount of granite.

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 74A

The heat lost is $1.56\times {10}^{8}J$

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:

$\begin{array}{l}m=3580kg\\ =3.58\times {10}^{6}g\\ \Delta T=41.2-(-12.9)\\ =54.1°C\\ c=0.803J/g-°C\end{array}$

Substituting the values in formula:

$\begin{array}{c}q=mc\Delta T\\ =3.58\times {10}^6\times 0.803\times 54.1\\ =1.56\times {10}^{8}J\end{array}$

Conclusion

Thus $1.56\times {10}^{8}J$heat is lost to decrease the temperature of given amount of granite.