Chapter 17.1, Problem 10LC

Interpretation Introduction

Interpretation: The heat absorbed by the water 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

  $\text{Q=m×C×ΔT}$

  $\text{Q}$ = 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 10LC

The heat absorbed by water will be ye $\text{1760 cal}$ or $7364\text{ Joules}$

Explanation of Solution

It is given that $32.0\text{ g}$ of water is heated from $\text{25}\text{.0°C}$ to $\text{80}\text{.0°C}$

  $\text{Q}$ = Heat absorbed by water = ?

  $\text{m}$ = mass of water = $32.0\text{ g}$

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

  $\text{ΔT}$ = change in temperature = $\text{80}\text{.0°C-25}\text{.0°C=55}\text{.0°C}$

Putting in the values we get,

  $\text{Q=32}\text{.0 g×1}\text{.0 cal/g°C×55}\text{.0°C=1760 cal}$

Also, the relation between calories and Joule is;

  $\begin{array}{l}\text{1}\text{.0 cal}=\text{4}\text{.184 Joules}\\ \text{1760 cal = }\frac{\text{4}\text{.184 Joules}}{\text{1}\text{.0 cal}}\times 1760\text{ cal}=7364\text{ Joules}\end{array}$