Chapter 15, Problem 77A

Interpretation Introduction

Interpretation:

The specific heat of canola oil is to be calculated for given amount 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 77A

The specific heat of canola oil is $1.91J/g-°C$

Explanation of Solution

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

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

Given data:

$\begin{array}{l}m=10.2g\\ \Delta T=196.4-25.0\\ =171.4°C\\ q=3.34kJ\\ =3340J\end{array}$

Substituting the values in formula:

$\begin{array}{c}c=\frac{q}{m\Delta T}\\ =\frac{3340}{10.2\times 171.4}\\ =1.91J/g-°C\end{array}$

Conclusion

Thus specific heat of canola oil is $1.91J/g-°C$.