Chapter 17, Problem 81A

Interpretation Introduction

Interpretation: The amount of water that could be heated from $25\text{°C}$ to the boiling point using a certain amount of energy is to be calculated.

Concept introduction: The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Mathematically it is represented by

  $\text{Q=m×C×ΔT}$

  ${\text{Q=m×C×(T}}_{\text{final}}-{\text{T}}_{\text{initial}})$

  $\text{Q}$ = Heat absorbed or released

  $\text{m}$ = mass of the substance

  $\text{C}$ = heat capacity of the substance

  $\text{ΔT}$ = change in temperature

Answer to Problem 81A

The amount of water that could be heated from $25\text{°C}$ to the boiling point using $\text{106 Calories}$ is $\text{1}\text{.41g}$

Explanation of Solution

The given values are

  $\text{Q}$ = heat absorbed by water = $\text{106 Calories}$

  $\text{m}$ = mass of water = ?

  ${\text{T}}_{\text{final}}$ = final temperature = boiling point of water = $100.0\text{°C}$

  ${\text{T}}_{\text{initial}}$ = initial temperature = $25.0\text{°C}$

  $\text{C}$ = specific heat of water = $\text{1}\text{.0 cal/g°C}$

Now put all the given values, we get

  $\begin{array}{l}106\text{ cal}=m\times 1.0\text{ cal/g°C}\times (100.0-25.0)\text{°C}\\ 106\text{ cal}=m\times 1.0\text{ cal/g°C}\times 75.0\text{°C}\\ \text{m=1}\text{.41g}\end{array}$