Chapter 17, Problem 96A

Interpretation Introduction

Interpretation: The mass of the ice cube 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 96A

The mass of the ice cube was $\text{77}\text{.8 g}$

Explanation of Solution

As we know that the heat released by one substance will be equal to the heat absorbed by the other substance.

  ${\text{m}}_{\text{1}}{\text{×C}}_{\text{1}}{\text{×(T}}_{\text{final}}{\text{-T}}_{\text{1}}{\text{)=-[m}}_{\text{2}}{\text{×C}}_{\text{2}}{\text{×(T}}_{\text{final}}{\text{-T}}_{\text{2}}\text{)]}$ $\text{(1)}$

where,

  ${\text{m}}_{\text{1}}$ = mass of ice = ?

  ${\text{m}}_{\text{2}}$ = mass of water = $30.0\text{ g}$

  ${\text{T}}_{\text{final}}$ = final temperature = $19.5\text{°C}$

  ${\text{T}}_{\text{1}}$ = initial temperature of ice = $0\text{°C}$

  ${\text{T}}_{\text{2}}$ = initial temperature of water = $45.0\text{°C}$

  ${\text{C}}_{\text{1}}$ = specific heat of ice = $\text{2}\text{.108J/g°C}$

  ${\text{C}}_{\text{2}}$ = specific heat of water = $\text{4}\text{.184J/g°C}$

Now put all the given values in equation $\text{(1)}$ , we get

  ${\text{m}}_{\text{1}}\text{×2}\text{.108J/g°C×(19}\text{.5°C-0°C)=-[30}\text{.0g×4}\text{.184J/g°C×(19}\text{.5-45}\text{.0°C)]}$

  $\begin{array}{l}{\text{m}}_1\text{×41}\text{.10J/g}=3200.76\text{J}\\ {\text{m}}_{\text{1}}\text{=77}\text{.8g}\end{array}$

Therefore, the mass of the ice cube was $\text{77}\text{.8 g}$