Chapter 7.2, Problem 7.3PP

Interpretation Introduction

Interpretation:

The temperature change has to be calculated, when $\text{50}\text{.0 g}$ of solution were in the calorimeter instead of $\text{1}{\text{.00×10}}^{\text{2}}\text{g}$ solution.

Concept Introduction:

Specific heat:

The required heat energy to rise a 1 degree Celsius temperature per gram of substance is known as specific heat.

    $\text{Q}\text{=}\text{S×m×ΔT}$

Where,

    $\text{ΔT}$ is temperature change,

    Q is heat,

  S is specific heat,

  m is mass.