Chapter 10, Problem 6STP

Interpretation Introduction

Interpretation: The heat change when 45.0 g of water cools from ${\text{45}}^{\text{0}}\text{C}$ to ${22}^{\text{0}}\text{C}$ needs to be determined.

Concept introduction: To calculate the energy required we use the following relation,

  $\text{Q}=s\times m\times \Delta T$

Here, Q = energy required

s = specific heat capacity

m = mass of the sample in grams

  $\Delta T$ = change in temperature in Celsius degrees

Answer to Problem 6STP

B is the correct option.

Explanation of Solution

Given,

specific heat capacity of water $\Delta T$ is $\text{4}\text{.184}{\text{J/g}}^{\text{0}}\text{C}$

The heat changes from ${\text{45}}^{\text{0}}\text{C}$ to ${22}^{\text{0}}\text{C}$

i.e.,

  $\begin{array}{c}\text{s=}\left({\text{45}}^{\text{0}}-{\text{22}}^{\text{0}}\right)\text{C }\\ \text{=}{\text{23}}^{\text{0}}\text{C}\end{array}$

Now. we calculate the energy required using the following formula.

  $\text{Q}=s\times m\times \Delta T$

  $\begin{array}{c}\text{Q=}\text{45g}\text{×}\text{23}\text{C}\text{×}\text{4}\text{.184}\text{J/g}\\ \text{=}\text{4}\text{.33}\text{kJ}\end{array}$

Conclusion

The heat change when 45.0 g of water cools from ${\text{45}}^{\text{0}}\text{C}$ to ${22}^{\text{0}}\text{C}$ is 4.33 kJ.

So, option A, C and D are incorrect.