Chapter 15, Problem 78A

Alloys When a 58.8-g piece of hot alloy is placed in125 g of cold water in a calorimeter, the temperature ofthe alloy decreases by 106.1°C, while the temperature ofthe water increases by 10.5°C. What is the specific heat ofthe alloy?

Interpretation Introduction

Interpretation:

Specific heatof alloy is to be described.

Concept introduction:

Specific heat: It is the amount of heat required to raise the temperature of an substance by ${\text{1}}_{}^{\text{o}}\text{C}$.

Answer to Problem 78A

Specific heat of the alloy is $\text{0}{\text{.880j/g}}_{}^{\text{o}}\text{C}$

Explanation of Solution

Since, $\text{Specific}\text{heat}\text{of}\text{alloy=}\text{-Specific}\text{heat}\text{of}\text{water}$

${\text{q}}_{\text{a}}^{}\text{=}{\text{-q}}_{\text{water}}^{}$

Relation between mass, temperature and specific heat.

${\text{m}}_{\text{a}}^{}{\text{(ΔT)}}_{\text{a}}^{}{\text{(C}}_{\text{a}}^{}\text{)=}{\text{-m}}_{\text{w}}^{}{\text{(ΔT)}}_{\text{w}}^{}{\text{(C}}_{\text{w}}^{}\text{)}$ where, m represents mass

C is the specific heat

$\text{ΔT}$ is the change in temperature

$\begin{array}{l}{\text{C}}_a^{}=-\frac{{\text{m}}_{\text{v}}^{}{\text{(ΔT)}}_{\text{w}}^{}{\text{(C}}_{\text{w}}^{}\text{)}}{{\text{m}}_{\text{a}}^{}{\text{(ΔT)}}_{\text{a}}^{}}\\ =\text{-}\frac{\text{(125)(10}\text{.5)(4}\text{.184)}}{\text{(58}\text{.8)(-106}\text{.1)}}\\ =\text{0}{\text{.880j/g}}_{}^{\text{o}}\text{C}\end{array}$

Conclusion

Since, heat transfers from one substance to the other, then q must be equal in magnitude, but opposite in direction for each substance.