Chapter 17.2, Problem 18LC

Interpretation Introduction

Interpretation: The heat released by the lead 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 18LC

The heat released by the lead is $502\text{J}$

Explanation of Solution

It is given that the lead is heated and placed in a calorimeter containing $40.0\text{ml}$ of water at $\text{17°C}$ and the temperature increases to $20.0\text{°C}$ .

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

  $\begin{array}{l}\text{Density of water = }\frac{\text{Mass of water}}{\text{Volume of water}}\\ \text{Mass of water}=\text{Density of water }\times \text{ Volume of water}\\ \text{Mass of water = 1g/ml}\times 40.0\text{ml = 40}\text{.0 g}\end{array}$

Putting in the values, we get:

  $Q$ = Heat absorbed by water = ?

  $m$ = mass of water = $\text{40}\text{.0 g}$

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

  $\Delta T$ = change in temperature = $(20.0\text{°C-17}\text{.0°C) = 3}\text{.00°C}$

Putting in the values we get,

  $\begin{array}{l}\text{Q}=40.0\text{g}\times 4.182\text{J/g°C}\times 3.00\text{°C}\\ \mathrm{Q}=502\text{J}\end{array}$