Chapter 15.2, Problem 13PP

Interpretation Introduction

Interpretation:

The mass of sample of water is to be calculated for given absorbed heat and rise in temperature.

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 13PP

The mass of sample of water is $50.77g$

Explanation of Solution

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

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

Given data:

$\begin{array}{l}c=4.184J/g-°C\\ \Delta T=46.6-20\\ =26.6°C\\ q=5650J\end{array}$

Substituting the values in formula:

$\begin{array}{c}m=\frac{q}{c\Delta T}\\ =\frac{5650}{4.184\times 26.6}\\ =50.77g\end{array}$

Conclusion

Thus mass of sample is $50.77g$