Chapter 10, Problem 52A

Interpretation Introduction

Interpretation: The amount of energy (in joules) required for heating $\text{2}\text{.5}\text{kg}$ of water from $18.5°\text{C}$ to $55.0°\text{C}$ should be calculated.

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.

Answer to Problem 52A

The amount of energy required to heat $\text{2}\text{.5}\text{kg}$ of water from $18.5°\text{C}$ to $55.0°\text{C}$ is $3.8\times {10}^5\text{J}$ .

Explanation of Solution

The amount of heat required for heating $\text{2}\text{.5}\text{kg}$ of water from $18.5°\text{C}$ to $55.0°\text{C}$ can be calculated using formula:

  $\text{q = mc}\Delta \text{T}$

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

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

Substituting the values:

  $\begin{array}{l}\text{q = 2}\text{.5 kg×}\frac{\text{1000 g}}{\text{1 kg}}\text{×4}\text{.184 J/g°C×}\left(\text{55}\text{.0-18}\text{.5}\right)\text{ °C}\\ \text{q = 10460 J/°C×}\left(\text{36}\text{.5}\right)\text{ °C}\\ \text{q = 3}{\text{.8×10}}^{\text{5}}\text{ J}\end{array}$

Hence, the amount of energy required to heat $\text{2}\text{.5}\text{kg}$ of water from $18.5°\text{C}$ to $55.0°\text{C}$ is $3.8\times {10}^5\text{J}$ .

Conclusion

  $3.8\times {10}^5\text{J}$ is the amount of energy required to heat $\text{2}\text{.5 kg}$ of water from $18.5°\text{C}$ to $55.0°\text{C}$ .