Chapter 10, Problem 54A

Interpretation Introduction

Interpretation: The value of final temperature of water needs to be determined if a sample of $\text{50}\text{.0 g}$ water at $100.0°\text{C}$ is poured into another sample of $\text{50}\text{.0 g}$ water at $25.0°\text{C}$ .

Concept introduction: Specific heat of capacity can be defined as the amount of heat added to a substance to raise the temperature by one degree. The expression for heat is:

  $\text{q = mc}\Delta \text{T}$ - (1)

Where q is heat, m is mass, c is specific heat and $\Delta \text{T}$ is change in temperature.

Answer to Problem 54A

The final temperature of water will be $\text{62}\text{.5 °C}$ .

Explanation of Solution

Initial temperature of both the water samples are $\text{100}\text{.0 °C}$ and $25.0°\text{C}$ . Mass of each water sample is $\text{50}\text{.0}\text{g}$ .

The specific heat capacity for the liquid water is $4.184\text{J/g°C}$ .

Let the final temperature be $T_2$ .

The heat lost by the water sample, which is at $\text{100}\text{.0 °C}$ can be determined by using equation (1) as:

  ${\text{q}}_{\text{lost}}\text{= 50}\text{.0 g×4}\text{.184 J/g°C×}\left({\text{T}}_{\text{2}}\text{-100}\text{.0}\right)\text{ °C}$

The heat gained by the water sample, which is at $25.0\text{ °C}$ can be determined by using equation (1) as:

  ${\text{q}}_{\text{gain}}\text{= 50}\text{.0 g×4}\text{.184 J/g°C×}\left({\text{T}}_{\text{2}}\text{-25}\text{.0}\right)\text{ °C}$

The heat lost by the one water sample will be equal to the heat gained by another water sample as:

  ${\text{q}}_{\text{gain}}{\text{= -q}}_{\text{lost}}$ (the negative sign represents that the heat is released during the process.)

Substituting the values in above expression as:

  $\begin{array}{l}\text{50}\text{.0 g×4}\text{.184 J/g°C×}\left({\text{T}}_{\text{2}}\text{-25}\text{.0}\right)\text{°C = - 50}\text{.0 g×4}\text{.184 J/g°C×}\left({\text{T}}_{\text{2}}\text{-100}\text{.0}\right)\text{°C}\\ \left({\text{T}}_{\text{2}}\text{-25}\text{.0}\right)\text{°C = -}\left({\text{T}}_{\text{2}}\text{-100}\text{.0}\right)\text{°C}\\ {\text{2T}}_{\text{2}}\text{= 125 °C}\\ {\text{T}}_{\text{2}}\text{= }\frac{\text{125 °C}}{2}\\ {\text{T}}_{\text{2}}\text{= 62}\text{.5 °C}\end{array}$

Hence, the final temperature of water will be $\text{62}\text{.5 °C}$ .