Chapter 17.2, Problem 13SP

Interpretation Introduction

Interpretation: The heat released by the pebble 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 13SP

The heat released by the pebble is $146.37\text{J}$

Explanation of Solution

It is given that the pebble is heated and placed in a calorimeter containing $25.0\text{ml}$ of water at $25\text{°C}$ and the temperature increases to $26.4\text{°C}$ .

The heat released by the pebble will be equal to the heat absorbed by the water in the calorimeter.

  $\begin{array}{l}\text{Density of water = }\frac{\text{mass of water}}{\text{Volume of water}}\\ \text{mass of water}=\text{Density of water }\times \text{ Volume of water}\\ \text{mass of water = 1g/ml }\times \text{ 25}\text{.0ml = 25}\text{.0g}\end{array}$

Putting in the values, we get:

  $Q$ = Heat absorbed by water= ?

  $m$ = mass of water = $\text{25}\text{.0g}$

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

  $\Delta T$ = change in temperature = $(26.4\text{°C-25°C) = 1}\text{.4°C}$

Putting in the values we get,

  $\begin{array}{l}Q=25.0\text{g}\times 4.182\text{J/g°C}\times 52\text{°C}\\ Q=146.37\text{J}\end{array}$