Chapter U5.98, Problem 8E

(a)

Interpretation Introduction

Interpretation:

The sample which will have higher temperature after 5 min needs to be determined from 50 g methanol and 50 g water kept in two 100 mL beaker placed on a hot plate.

Concept Introduction:

Heat absorbed by any sample H $\text{=m×s×t}$ where m, s, t are mass, specific heat and temperature rise of the sample.

Answer to Problem 8E

Methanol sample will have higher temperature.

Explanation of Solution

Specific heat (s) of water is greater than that of methanol.

So, when mass of the water and methanol sample (m) is same and same heat (H) is provided by the hot plate then methanol will have more temperature rise as it has lesser specific heat (s).

b)

Interpretation Introduction

Interpretation:

Final temperature for methanol and water must be found when initial temperature is 23°C and 25 cal heat energy is provided by the hot plate.

Concept Introduction:

Relation of heat (H) and temperature change (t) can be given by $\text{H=m×s×t}$ where m and s are mass and specific heat of the sample.

Answer to Problem 8E

Temperature of water will be 23.5°C and temperature of methanol will be 23.83°C.

Explanation of Solution

For water

  $\begin{array}{l}\text{25 cal=50 g×1 }\frac{\text{cal}}{\text{g/}°C}\text{× t}\\ \text{Or, t=}\frac{25}{50}°C=0.5°C.\end{array}$

So final temperature $\text{=(23+0}\text{.5)°C=23}\text{.5°C}\text{.}$

For methanol,

  $\begin{array}{l}\text{25 cal=50 g×0}\text{.6 }\frac{\text{cal}}{\text{g/}°C}\text{× t}\\ \text{Or, t=}\frac{25}{50\times 0.6}°C=\frac{25}{30}°C=0.83°C.\end{array}$

So final temperature $\text{=(23+0}\text{.83)°C=23}\text{.83°C}\text{.}$