Chapter 3.5, Problem 39PP

(a)

Interpretation Introduction

Interpretation : The specific heat of $\text{13}\text{.0 g}$ piece of glass must be calculated if it is dropped into $\text{38 g}$ of water at $\text{22}\text{.9 }°\text{C}$ .The final temperature of water and glass is $\text{25}\text{.6}°\text{C}$ .

Concept Introduction : The temperature change can be used along with the specific heat and mass of the solution to calculate the amount of heat involved or released during the process. The heat change and specific heat can be related with the given formula:

  $\text{q = m }\times \text{ C }\times \Delta \text{T}$

Here;

• m = mass of substance
• C = specific heat
• $\Delta \text{T}$ = Temperature change

(b)

Interpretation Introduction

Interpretation : The specific heat of $\text{19}\text{.6 g}$ piece of metalshould be calculated if it is dropped into $\text{42 g}$ of water at $\text{24}\text{.2}°\text{C}$ . The final temperature of water and metal is $\text{27}\text{.6 }°\text{C}$ .

Concept Introduction : The temperature change can be used along with the specific heat and mass of the solution to calculate the amount of heat involved or released during the process. The heat change and specific heat can be related with the given formula:

  $\text{q = m }\times \text{ C }\times \Delta \text{T}$

Here;

• m = mass of substance
• C = specific heat
• $\Delta \text{T}$ = Temperature change