Chapter 15.2, Problem 15PP

Interpretation Introduction

Interpretation:

The final temperature of water is to be calculated when water loses heat.

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 15PP

The final temperature is $18.04°C$

Explanation of Solution

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

$\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatlost\\ m=givenmass\\ T_1=initialtemperature\\ T_2=finaltemperature\\ \Delta T=fallintemperature\\ cisspecificheat\\ \therefore \Delta T=\frac{q}{mc}\\ or\\ T_1-T_2=\frac{q}{mc}(asheatislost)\end{array}$

Given data:

$\begin{array}{l}m=335g\\ T_1=65.5°C\\ q=9750J\\ c=4.184J/g-°C\end{array}$

Substituting the values in formula:

$\begin{array}{c}{T}_1-T_2=\frac{q}{mc}\\ 25-T_2=\frac{9750}{335\times 4.184}\\ 25-T_2=6.96°C\\ T_2=18.04°C\end{array}$

Conclusion

Thus, the final temperature of water is $18.04°C$