Chapter 7, Problem 1E

Calculate the quantity of heat, in kilojoules, (a) required to raise the temperature of 9.25 L of water from 22.0 to 29.4 °C; (b) associated with a 33.5°C decrease in temperature in a 5.85 kg aluminum bar (specific heat capacity of aluminum $\text{= 0}{\text{.903 J g}}^{\text{-10}}{\text{C}}^{\text{-1}}\text{)}\text{.}$

Interpretation Introduction

(a)

Interpretation:

The heat required to raise the temperature of 9.25 L water from 22 ⁰C to 29.4 ⁰C is to be calculated.

Concept introduction:

The amount of heat required in order to increase the temperature by one degree Celsius of unit mass of a substance is said to be the specific heat.

The heat required to change the temperature of a substance is related to heat capacity of the substance by the expression as:

$\begin{array}{l}Q=mc\Delta T\\ \text{where,}\\ \text{Q = Heat required}\\ \text{m = mass of the substance}\\ \text{c = specific heat of the substance}\\ \Delta \text{T = change in temperature}\end{array}$

Answer to Problem 1E

The heat required to raise the temperature of 9.25 L water from 22 ⁰C to 29.4 ⁰C is 286.121 kJ.

Explanation of Solution

The amount of heat required to raise the temperature of a substance is given as:

$\begin{array}{l}Q=mc\Delta T----(1)\\ \text{where,}\\ \text{Q = Heat required}\\ \text{m = mass of the substance}\\ \text{c = specific heat of the substance}\\ \Delta \text{T = change in temperature}\end{array}$

The mass of water can be calculated using the given volume of water and the density of water.

$\begin{array}{l}\text{Density of water = }\frac{\text{Mass of water}}{\text{Volume of water }}\\ \Rightarrow \text{ Mass of water = Density of water }\times \text{ Volume of water}\\ \Rightarrow \text{m}=1\text{ kg/L }\times \text{ 9}\text{.25 L = 9}\text{.25 kg}\\ \Rightarrow \text{m}=9.25\times 1000\text{ g (}\because \text{ 1 kg = 1000 g)}\\ \Rightarrow \text{m = 9250 g}\end{array}$

The change in temperature = (29.4 - 22.0) ⁰C = 7.4 ⁰C

Specific heat of water = c = 4.18 J/g⁰C

Substituting the above values in equation (1) to calculate the value of Q:

$\begin{array}{l}Q=mc\Delta T\\ \Rightarrow Q=9250\text{ g }\times \text{ 4}{\text{.18 J/g}}^{\text{0}}\text{C }\times \text{ (29}{\text{.4 - 22) }}^{\text{0}}\text{C}\\ \Rightarrow Q=286121\text{ J}\\ \Rightarrow Q=286121\times {10}^{-3}\text{ kJ}\\ \Rightarrow Q=286.121\text{ kJ}\end{array}$

Therefore, the heat required to raise the temperature of 9.25 L of water is 286.121 kJ.

Interpretation Introduction

(b)

Interpretation:

The quantity of heat associated with a 33.5 ⁰C decrease in temperature in a 5.85 kg aluminum bar is to be calculated.

Concept introduction:

The amount of heat required in order to increase the temperature by one degree Celsius of unit mass of a substance is said to be the specific heat.

The heat required to change the temperature of a substance is related to heat capacity of the substance by the expression as:

$\begin{array}{l}Q=mc\Delta T\\ \text{where,}\\ \text{Q = Heat required}\\ \text{m = mass of the substance}\\ \text{c = specific heat of the substance}\\ \Delta \text{T = change in temperature}\end{array}$

Answer to Problem 1E

The quantity of heat required to decrease the temperature of aluminum bar by 33.5 ⁰C is 176.97 kJ.

Explanation of Solution

The amount of heat required to change the temperature of a substance is given as:

$\begin{array}{l}Q=mc\Delta T----(1)\\ \text{where,}\\ \text{Q = Heat required}\\ \text{m = mass of the substance}\\ \text{c = specific heat of the substance}\\ \Delta \text{T = change in temperature}\end{array}$

The mass of aluminum is given to be

$\text{m = 5}\text{.85 kg = 5}\text{.85}\times {\text{10}}^3\text{ g = 5850 g}$

The decrease in temperature = 33.5 ⁰C

Specific heat of aluminum = c = 0.903 J/g⁰C

Substituting the above values in equation (1) to calculate the value of Q:

$\begin{array}{l}Q=mc\Delta T\\ \Rightarrow Q=5850\text{ g }\times \text{ 0}{\text{.903 J/g}}^{\text{0}}\text{C }\times \text{ 33}{\text{.5 }}^{\text{0}}\text{C}\\ \Rightarrow Q=176965.425\text{ J}\\ \Rightarrow Q=176965.425\times {10}^{-3}\text{ kJ}\\ \Rightarrow Q=176.97\text{ kJ}\end{array}$

Therefore, the heat required to decrease the temperature of 5.85 kg aluminum by 33.5 ⁰C is 176.97 kJ.