Chapter 17, Problem 69A

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

  $Q=m\times C\times \Delta T$

  $Q$ = Heat absorbed or released

  $m$ = mass of the substance

  $C$ = heat capacity of the substance

  $\Delta T$ = change in temperature

Answer to Problem 69A

The heat absorbed by $1.00\text{ L}$ of water when it is heated from $18.0\text{°C}$ to $85.0\text{°C}$ is $\text{280}\text{.194 kJ}$

Explanation of Solution

It is given that $1.00\text{ L}$ of water when it is heated from $18.0\text{°C}$ to $85.0\text{°C}$

  $\text{Volume of water = 1}\text{.00 L= 1}\text{.00times 1000 ml=1000 ml}$

  $\begin{array}{l}\text{Density of water = }\frac{\text{Mass of water}}{\text{Volume of water}}\\ \text{Mass of water}=\text{Density of water }\times \text{ Volume of water}\\ \text{Mass of water = 1 g/ml}\times 1000\text{ml = 1000 g}\end{array}$

Putting in the values, we get:

  $Q$ = Heat absorbed by water = ?

  $m$ = mass of water = $\text{1000 g}$

  $C$ = heat capacity of water = $\text{4}\text{.182 J/g°C}$

  $\Delta T$ = change in temperature = $(85.0\text{°C-18}\text{.0°C})=67\text{°C}$

Putting in the values we get,

  $\begin{array}{l}\text{Q}=1000\text{g}\times 4.182\text{J/g°C}\times 67\text{°C}\\ \mathrm{Q}=280194\text{J=280}\text{.194 kJ}\end{array}$